Prove that t a n A + s e c A − 1 t a n A − s e c A + 1 = 1 + S i n A c o s A .

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: L . H . S . t a n A + s e c A − 1 t a n A − s e c A + 1 = t a n A + ( s e c A − 1 ) t a n A − ( s e c A − 1 ) = [ t a n A + ( s e c A − 1 ) ] [ t a n A − ( s e c A − 1 ) ] × [ t a n A + ( s e c A − 1 ) ] [ t a n A − ( s e c A − 1 ) ] [ Rationalizing the denominator ] = [ t a n A + ( s e c A − 1 ) ] 2 t a n 2 A − ( s e c A − 1 ) 2 = t a n 2 A + ( s e c A − 1 ) 2 + 2 t a n A ( s e c A − 1 ) t a n 2 A − ( s e c 2 A + 1 − 2 s e c A ) = t a n 2 A + s e c 2 A + 1 − 2 s e c A + 2 t a n A s e c A − 2 t a n A t a n 2 A − s e c 2 A − 1 + 2 s e c A = s e c 2 A + s e c 2 A − 2 s e c A + 2 t a n A s e c A − 2 t a n A − 1 − 1 + 2 s e c A = 2 s e c 2 A − 2 s e c A + 2 t a n A s e c A − 2 t a n A 2 s e c A − 2 = s e c 2 A − s e c A + t a n A s e c A − t a n A s e c A − 1 = s e c A ( s e c A − 1 ) + t a n A ( s e c A − 1 ) s e c A − 1 = ( s e c A − 1 ) ( s e c A + t a n A ) ( s e c A − 1 ) = s e c A + t a n A = 1 c o s A + s i n A c o s A = 1 + s i n A c o s A R . H . S . H e n c e p r o v e d .

If 2 s i n α 1 + c o s α + s i n α , = 1 then prove that 1 − c o s α + s i n α 1 + s i n α = i s a l s o e q u a l t o y . [Hint: Express 1 − c o s α + s i n α 1 + s i n α = 1 − c o s α + s i n α 1 + s i n α = 1 + c o s α + s i n α 1 + c o s α + s i n α

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : y = 2 s i n α 1 + c o s α + s i n α = 2 s i n α 1 + c o s α + s i n α × 1 + s i n α − c o s α 1 + s i n α − c o s α = 2 s i n α ( 1 − c o s α + s i n α ) ( 1 + s i n α ) 2 − c o s 2 α = 2 s i n α ( 1 − c o s α + s i n α ) 1 + s i n 2 α + 2 s i n α − c o s 2 α = 2 s i n α ( 1 − c o s α + s i n α ) ( 1 − c o s 2 α ) + s i n 2 α + 2 s i n α = 2 s i n α ( 1 − c o s α + s i n α ) s i n 2 α + s i n 2 α + 2 s i n α = 2 s i n α ( 1 − c o s α + s i n α ) 2 s i n 2 α + 2 s i n α = 2 s i n α ( 1 − c o s α + s i n α ) 2 s i n α ( 1 + s i n α ) = 1 − c o s α + s i n α 1 + s i n α = y H e n c e p r o v e d .

If m s i n θ = n s i n ( θ + 2 α ) , then prove that t a n ( θ + α ) c o t α = m + n m − n . [Hint: Express s i n ( θ + 2 α ) s i n θ and apply componendo and dividendo.]

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : m s i n θ = n s i n ( θ + 2 α ) ⇒ s i n ( θ + 2 α ) s i n θ = m n Using componendo and dividendo theorm we get ⇒ s i n ( θ + 2 α ) + s i n θ s i n ( θ + 2 α ) − s i n θ = m + n m − n ⇒ 2 s i n ( θ + 2 α + θ 2 ) . c o s ( θ + 2 α − θ 2 ) 2 c o s ( θ + 2 α + θ 2 ) . s i n ( θ + 2 α − θ 2 ) = m + n m − n [ ? s i n A + s i n B = 2 s i n A + B 2 . c o s A − B 2 s i n A − s i n B = 2 c o s A + B 2 . s i n A − B 2 ] ⇒ s i n ( θ + α ) . c o s α c o s ( θ + α ) − s i n α = m + n m − n ⇒ t a n ( θ + α ) . c o t α = m + n m − n H e n c e p r o v e d .

If c o s ( α + β ) = 4 5 and s i n ( α − β ) = 5 1 3 , where α lies between 0 and π 4 , find the value of t a n 2 α . [Hint: Express t a n 2 α as t a n ( α + β + α − β ) .]

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : c o s ( α + β ) = 4 5 ∴ t a n ( α + β ) = 3 4 a n d s i n ( α − β ) = 5 1 3 ∴ t a n ( α − β ) = 5 1 2 N o w t a n 2 α = t a n [ α + β + α − β ] = t a n [ ( α + β ) + ( α − β ) ] = t a n ( α + β ) + t a n ( α − β ) 1 − t a n ( α + β ) . t a n ( α − β ) = 3 4 + 5 1 2 1 − 3 4 × 5 1 2 = 9 + 5 1 2 4 8 − 1 5 4 8 = 1 4 1 2 × 4 8 3 3 = 5 6 3 3 H e n c e , t a n 2 α = 5 6 3 3 .

If t a n x = b a , then find the value of a + b a − b + a − b a + b .

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : t a n x = b a a + b a − b + a − b a + b = a + b a − b + a − b a + b = a + b + a − b ( a − b ) ( a + b ) = 2 a a 2 − b 2 = 2 a a 1 − b 2 a 2 = 2 1 − t a n 2 x [ ? t a n x = b a ] = 2 1 − s i n 2 x c o s 2 x = 2 c o s 2 x − s i n 2 x c o s x = 2 c o s x c o s 2 x [ ? c o s 2 x = c o s 2 x − s i n 2 x ] H e n c e , a + b a − b + a − b a + b = 2 c o s x c o s 2 x

The value of s i n 5 0 ∘ s i n 1 3 0 ∘ is ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar Sol: s i n 5 0 0 s i n 1 3 0 0 = s i n 5 0 0 s i n ( 1 8 0 0 − 5 0 0 ) = s i n 5 0 0 s i n 5 0 0 = 1 H e n c e , t h e v a l u e o f f i l l e r i s 1 .

If k = s i n π 1 8 s i n 5 π 1 8 s i n 7 π 1 8 , then the numerical value of k is ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : k = s i n ( π 1 8 ) s i n ( 5 π 1 8 ) s i n ( 7 π 1 8 ) ⇒ k = s i n 1 0 0 . s i n 5 0 0 . s i n 7 0 0 ⇒ k = s i n 1 0 0 . s i n ( 9 0 0 − 4 0 0 ) . s i n ( 9 0 0 − 2 0 0 ) ⇒ k = s i n 1 0 0 . c o s 4 0 0 . c o s 2 0 0 ⇒ k = s i n 1 0 0 . 1 2 [ 2 c o s 4 0 0 . c o s 2 0 0 ] ⇒ k = s i n 1 0 0 . 1 2 [ c o s ( 4 0 0 + 2 0 0 ) + c o s ( 4 0 0 − 2 0 0 ) ] ⇒ k = 1 2 s i n 1 0 0 [ c o s 6 0 0 + c o s 2 0 0 ] ⇒ k = 1 2 s i n 1 0 0 [ 1 2 + c o s 2 0 0 ] ⇒ k = 1 4 s i n 1 0 0 + 1 2 s i n 1 0 0 c o s 2 0 0 ⇒ k = 1 4 s i n 1 0 0 + 1 4 ( 2 s i n 1 0 0 c o s 2 0 0 ) ⇒ k = 1 4 s i n 1 0 0 + 1 4 [ s i n ( 1 0 0 + 2 0 0 ) + s i n ( 1 0 0 − 2 0 0 ) ] ⇒ k = 1 4 s i n 1 0 0 + 1 4 [ s i n 3 0 0 + s i n ( − 1 0 0 ) ] ⇒ k = 1 4 s i n 1 0 0 + 1 4 s i n 3 0 0 − 1 4 s i n 1 0 0 ⇒ k = 1 4 s i n 3 0 0 = 1 4 × 1 2 = 1 8 . H e n c e , t h e v a l u e o f f i l l e r i s 1 8 .

If t a n A = 1 − c o s B s i n B , then t a n 2 A is ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : t a n A = 1 − c o s B s i n B t a n 2 A = 2 t a n A 1 − t a n 2 A = 2 ( 1 − c o s B s i n B ) 1 − ( 1 − c o s B s i n B ) 2 = 2 ( 2 s i n 2 B / 2 2 s i n B / 2 c o s B / 2 ) 1 − ( 2 s i n 2 B / 2 2 s i n B / 2 c o s B / 2 ) 2 [ ? 1 − c o s B = 2 s i n 2 B / 2 s i n B = 2 s i n B / 2 c o s B / 2 ] = 2 ( s i n B / 2 c o s B / 2 ) 1 − ( s i n B / 2 c o s B / 2 ) 2 = 2 t a n B / 2 1 − t a n 2 B / 2 = t a n B S o , t a n 2 A = t a n B H e n c e , t h e v a l u e o f f i l l e r i s t a n B .

If c o s ( θ + ϕ ) = m c o s ( θ − ϕ ) , then prove that tan θ = 1 − m 1 + m . c o t ϕ [Hint: Express c o s ( θ + α ) c o s ( θ − α ) = m 1 .]and apply componendo and dividendo.]

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : c o s ( θ + ? ) = m c o s ( θ − ? ) ⇒ c o s ( θ + ? ) c o s ( θ − ? ) = m 1 Using componendo and dividendo theorem, we get c o s ( θ + ? ) + c o s ( θ − ? ) c o s ( θ + ? ) − c o s ( θ − ? ) = m + 1 m − 1 ⇒ 2 c o s ( θ + ? + θ − ? 2 ) . c o s ( θ + ? − θ + ? 2 ) − 2 s i n ( θ + ? + θ − ? 2 ) . s i n ( θ + ? − θ + ? 2 ) = m + 1 m − 1 ⇒ c o s θ . c o s ? − s i n θ . s i n ? = m + 1 m − 1 ⇒ − c o t θ . c o t ? = m + 1 m − 1 ⇒ − c o t ? t a n θ = m + 1 m − 1 = − 1 + m 1 − m ⇒ t a n θ = 1 + m 1 − m c o t ? . H e n c e p r o v e d .

Find the value of the expression 3 [ s i n 4 ( 3 π 2 − α ) + s i n 4 ( 3 π + α ) ] − 2 [ s i n 6 ( 2 π / 2 + α ) + s i n 6 ( 5 π − α ) ] .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : 3 [ s i n 4 ( 3 π 2 − α ) + s i n 4 ( 3 π + α ) ] − 2 [ s i n 6 ( π 2 + α ) + s i n 6 ( 5 π − α ) ] = 3 [ c o s 4 α + s i n 4 ( π + α ) ] − 2 [ c o s 6 α + s i n 6 ( π − α ) ] = 3 [ c o s 4 α + s i n 4 α ] − 2 [ c o s 6 α + s i n 6 α ] = 3 [ c o s 4 α + s i n 4 α + 2 s i n 2 α c o s 2 α − 2 s i n 2 α c o s 2 α ] − 2 [ ( c o s 2 α + s i n 2 α ) 3 − 3 c o s 2 α s i n 2 α ( c o s 2 α + s i n 2 α ) ] = 3 [ ( c o s 2 α + s i n 2 α ) 2 − 2 s i n 2 α c o s 2 α ] − 2 [ 1 − 3 c o s 2 α s i n 2 α ] = 3 [ 1 − 2 s i n 2 α c o s 2 α ] − 2 [ 1 − 3 c o s 2 α s i n 2 α ] = 3 − 6 s i n 2 α c o s 2 α − 2 + 6 c o s 2 α s i n 2 α = 3 − 2 = 1 Hence, the value of the given expression is 1.

If a c o s 2 θ + b s i n 2 θ = c , has α and β as its roots, then prove that t a n α + t a n β = 2 b a + c . [Hint: Use the identities for c o s 2 θ = 1 − t a n 2 θ 1 + t a n 2 θ and s i n 2 θ = 2 t a n θ 1 + t a n 2 θ ]

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : a c o s 2 θ + b s i n 2 θ = c ⇒ a [ 1 − t a n 2 θ 1 + t a n 2 θ ] + b [ 2 t a n θ 1 + t a n 2 θ ] = c ⇒ a − a t a n 2 θ + 2 b t a n θ = c ( 1 + t a n 2 θ ) [ ? c o s 2 θ = 1 − t a n 2 θ 1 + t a n 2 θ , s i n 2 θ = 2 t a n θ 1 + t a n 2 θ ] ⇒ a − a t a n 2 θ + 2 b t a n θ − c t a n 2 θ − c = 0 ⇒ − ( a + c ) t a n 2 θ + 2 b t a n θ + ( a − c ) = 0 ⇒ ( a + c ) t a n 2 θ − 2 b t a n θ + ( c − a ) = 0 Since α and β are the roots of this equation ⇒ t a n α + t a n β = − ( − 2 b ) a + c ⇒ t a n α + t a n β = 2 b a + c . H e n c e p r o v e d .

If f ( x ) = c o s 2 x + s e c 2 x , then: (a) f ( x ) < 1 (b) f ( x ) = 1 (c) 2 < f ( x ) < 1 (d) f(x) ≥2 [Hint: A.M ≥ G.M.]

This is a Objective Type Questions as classified in NCERT Exemplar Sol: Given that:f(x)=cos2x+sec2x W e k n o w t h a t A M ≥ G M ⇒ cos2x+sec2x 2 ≥ cos2x.sec2x ⇒ cos2x+sec2x2≥1 ⇒cos2x+sec2x≥2 ⇒ f ( x ) ≥ 2 H e n c e , t h e c o r r e c t o p t i o n i s ( d ) .

If t a n θ = 1 2 = and t a n ϕ = 1 / 3 , then the value of θ + ϕ is: (a) π 6 (b) π (c) 0 (d) π 4

This is an Objective Type Questions as classified in NCERT Exemplar Sol: W e k n o w t h a t t a n ( θ + ? ) = t a n θ + t a n ? 1 − t a n θ t a n ? = 1 2 + 1 3 1 − 1 2 × 1 3 = 5 6 5 6 = 1 ⇒ t a n ( θ + ? ) = t a n π 4 ∴ θ + ? = π 4 . H e n c e , t h e c o r r e c t o p t i o n i s ( d ) .

Which of the following is not correct? (a) s i n θ = − 1 5 (b) c o s θ = 1 (c) s e c θ = 2 (d) t a n θ = 2 0

This is an Objective Type Questions as classified in NCERT Exempar Sol: sinθ=−15 is correct. ?−1≤sinθ≤1So (a) is correct.cosθ=1 is correct. ?cos00=1So (b) is correct.secθ=12 ⇒cosθ=2 is not correct. ?−1≤cosθ≤1Hence, the correct option is (c). s i n θ = − 1 5 i s c o r r e c t . ? − 1 ≤ s i n θ ≤ 1 S o ( a ) i s c o r r e c t . c o s θ = 1 i s c o r r e c t . ? c o s 0 0 = 1 S o ( b ) i s c o r r e c t . s e c θ = 1 2 ⇒ c o s θ = 2 i s n o t c o r r e c t . ? − 1 ≤ c o s θ ≤ 1 H e n c e , t h e c o r r e c t o p t i o n i s ( c ) .

The value of t a n 1 ∘ ⋅ t a n 2 ∘ ⋅ t a n 3 ∘ ⋯ t a n 8 9 ∘ is: (a) 0 (b) 1 (c) 2 (d) Not defined

This is an Objective Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : t a n 1 0 t a n 2 0 t a n 3 0 … t a n 8 9 0 = t a n 1 0 t a n 2 0 t a n 3 0 … t a n 4 5 0 . t a n ( 9 0 − 4 4 ) 0 . t a n ( 9 0 − 4 3 ) 0 … t a n ( 9 0 − 1 ) 0 = t a n 1 0 c o t 1 0 . t a n 2 0 c o t 2 0 . t a n 3 0 c o t 3 0 … t a n 8 9 0 . c o t 8 9 0 = 1 . 1 . 1 . 1 … 1 . 1 = 1 H e n c e , t h e c o r r e c t o p t i o n i s ( b ) .

If t a n A = 1 − c o s B s i n B , then t a n 2 A = t a n B .

This is an True or False Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : t a n A = 1 − c o s B s i n B = 2 s i n 2 B / 2 2 s i n B / 2 c o s B / 2 = t a n B 2 t a n 2 A = 2 t a n A 1 − t a n 2 A = 2 t a n B / 2 1 − t a n 2 B / 2 ∴ t a n 2 A = t a n B H e n c e , t h e s t a t e m e n t i s ' T r u e ' .

s i n 1 0 ∘ is greater than c o s 1 0 ∘ .

This is an True or False Type Questions as classified in NCERT Exemplar Sol: I f s i n 1 0 0 > c o s 1 0 0 ⇒ s i n 1 0 0 > c o s ( 9 0 0 − 8 0 0 ) ⇒ s i n 1 0 0 > s i n 8 0 0 w h i c h i s n o t p o s s i b l e . H e n c e , t h e s t a t e m e n t i s ' F a l s e ' .

. c o s π 1 5 ⋅ c o s 4 π 1 5 ⋅ c o s 8 π 1 5 ⋅ c o s 1 6 π 1 5 = 1 1 6

This is an True or False Type Questions as classified in NCERT Exemplar Sol: L . H . S . c o s 2 π 1 5 . c o s 4 π 1 5 . c o s 8 π 1 5 . c o s 1 6 π 1 5 = c o s 2 4 0 . c o s 4 8 0 . c o s 9 6 0 . c o s 1 9 2 0 = 1 1 6 s i n 2 4 0 [ ( 2 s i n 2 4 0 c o s 2 4 0 ) ( 2 c o s 4 8 0 ) ( 2 c o s 9 6 0 ) ( 2 c o s 1 9 2 0 ) ] = 1 1 6 s i n 2 4 0 [ s i n 4 8 0 . 2 c o s 4 8 0 ( 2 c o s 9 6 0 ) ( 2 c o s 1 9 2 0 ) ] = 1 1 6 s i n 2 4 0 [ 2 s i n 4 8 0 c o s 4 8 0 ( 2 c o s 9 6 0 ) ( 2 c o s 1 9 2 0 ) ] = 1 1 6 s i n 2 4 0 [ s i n 9 6 0 ( 2 c o s 9 6 0 ) ( 2 c o s 1 9 2 0 ) ] = 1 1 6 s i n 2 4 0 [ 2 s i n 9 6 0 c o s 9 6 0 ( 2 c o s 1 9 2 0 ) ] = 1 1 6 s i n 2 4 0 [ s i n 1 9 2 0 ( 2 c o s 1 9 2 0 ) ] = 1 1 6 s i n 2 4 0 [ 2 s i n 1 9 2 0 c o s 1 9 2 0 ] = 1 1 6 s i n 2 4 0 s i n 3 8 4 0 = 1 1 6 s i n 2 4 0 s i n ( 3 6 0 0 + 2 4 0 ) = 1 1 6 s i n 2 4 0 × s i n 2 4 0 [ ? s i n ( 3 6 0 0 + θ ) = s i n θ ] = 1 1 6 R . H . S . H e n c e , t h e s t a t e m e n t i s ' T r u e ' .

Maths NCERT Exemplar Solutions Class 11th Chapter Three: Overview, Questions, Preparation

Updated on Aug 29, 2025 13:59 IST

By alok kumar singh, Executive Content Operations

Table of contents

Trigonometric Functions Short Answer Type Questions
Trigonometric Functions Long Answer Type Questions
Trigonometric Functions Objective Type Questions
Trigonometric Functions Fill in the blanks Type Questions
Trigonometric Functions True or False Type Questions
JEE Mains 2020

Trigonometric Functions Short Answer Type Questions

1. Prove that $\frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} = \frac{1 + S i n A}{c o s A} .$

Sol:

$\begin{array}{l} L . H . S . \frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} \\ = \frac{t a n A + (s e c A - 1)}{t a n A - (s e c A - 1)} \\ = \frac{[t a n A + (s e c A - 1)]}{[t a n A - (s e c A - 1)]} \times \frac{[t a n A + (s e c A - 1)]}{[t a n A - (s e c A - 1)]} [R a t i o n a l i z i n g t h e] \\ = \frac{{[t a n A + (s e c A - 1)]}^{2}}{t a n^{2} A - {(s e c A - 1)}^{2}} \\ = \frac{t a n^{2} A + {(s e c A - 1)}^{2} + 2 t a n A (s e c A - 1)}{t a n^{2} A - (s e c^{2} A + 1 - 2 s e c A)} \\ = \frac{t a n^{2} A + s e c^{2} A + 1 - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{t a n^{2} A - s e c^{2} A - 1 + 2 s e c A} \\ = \frac{s e c^{2} A + s e c^{2} A - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{- 1 - 1 + 2 s e c A} \\ = \frac{2 s e c^{2} A - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{2 s e c A - 2} \\ = \frac{s e c^{2} A - s e c A + t a n A s e c A - t a n A}{s e c A - 1} \\ = \frac{s e c A (s e c A - 1) + t a n A (s e c A - 1)}{s e c A - 1} \\ = \frac{(s e c A - 1) (s e c A + t a n A)}{(s e c A - 1)} \\ = s e c A + t a n A = \frac{1}{c o s A} + \frac{s i n A}{c o s A} = \frac{1 + s i n A}{c o s A} R . H . S . H e n c e p r o v e d . \end{array}$

2. If $\frac{2 s i n α}{1 + c o s α + s i n α}, = 1$ then prove that $\frac{1 - c o s α + s i n α}{1 + s i n α} = i s a l s o e q u a l t o y .$

[Hint: Express $\frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 + c o s α + s i n α}{1 + c o s α + s i n α}$

Sol:

$\begin{array}{l} G i v e n t h a t : \\ y = \frac{2 s i n α}{1 + c o s α + s i n α} \\ = \frac{2 s i n α}{1 + c o s α + s i n α} \times \frac{1 + s i n α - c o s α}{1 + s i n α - c o s α} \\ = \frac{2 s i n α (1 - c o s α + s i n α)}{{(1 + s i n α)}^{2} - c o s^{2} α} = \frac{2 s i n α (1 - c o s α + s i n α)}{1 + s i n^{2} α + 2 s i n α - c o s^{2} α} \\ = \frac{2 s i n α (1 - c o s α + s i n α)}{(1 - c o s^{2} α) + s i n^{2} α + 2 s i n α} = \frac{2 s i n α (1 - c o s α + s i n α)}{s i n^{2} α + s i n^{2} α + 2 s i n α} \\ = \frac{2 s i n α (1 - c o s α + s i n α)}{2 s i n^{2} α + 2 s i n α} = \frac{2 s i n α (1 - c o s α + s i n α)}{2 s i n α (1 + s i n α)} \\ = \frac{1 - c o s α + s i n α}{1 + s i n α} = y \\ H e n c e p r o v e d . \end{array}$

Commonly asked questions

Prove that $\frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} = \frac{1 + S i n A}{c o s A} .$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L . H . S . \frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} \\ = \frac{t a n A + (s e c A - 1)}{t a n A - (s e c A - 1)} \\ = \frac{[t a n A + (s e c A - 1)]}{[t a n A - (s e c A - 1)]} \times \frac{[t a n A + (s e c A - 1)]}{[t a n A - (s e c A - 1)]} [R a t i o n a l i z i n g t h e denominator] \\ = \frac{{[t a n A + (s e c A - 1)]}^{2}}{t a n^{2} A - {(s e c A - 1)}^{2}} \\ = \frac{t a n^{2} A + {(s e c A - 1)}^{2} + 2 t a n A (s e c A - 1)}{t a n^{2} A - (s e c^{2} A + 1 - 2 s e c A)} \\ = \frac{t a n^{2} A + s e c^{2} A + 1 - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{t a n^{2} A - s e c^{2} A - 1 + 2 s e c A} \\ = \frac{s e c^{2} A + s e c^{2} A - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{- 1 - 1 + 2 s e c A} \\ = \frac{2 s e c^{2} A - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{2 s e c A - 2} \\ = \frac{s e c^{2} A - s e c A + t a n A s e c A - t a n A}{s e c A - 1} \\ = \frac{s e c A (s e c A - 1) + t a n A (s e c A - 1)}{s e c A - 1} \\ = \frac{(s e c A - 1) (s e c A + t a n A)}{(s e c A - 1)} \\ = s e c A + t a n A = \frac{1}{c o s A} + \frac{s i n A}{c o s A} = \frac{1 + s i n A}{c o s A} R . H . S . H e n c e p r o v e d . \end{array}$

If $\frac{2 s i n α}{1 + c o s α + s i n α}, = 1$ then prove that $\frac{1 - c o s α + s i n α}{1 + s i n α} = i s a l s o e q u a l t o y .$

[Hint: Express $\frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 + c o s α + s i n α}{1 + c o s α + s i n α}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

If $m s i n θ = n s i n (θ + 2 α)$ , then prove that $t a n (θ + α) c o t α = \frac{m + n}{m - n} .$

[Hint: Express $\frac{s i n (θ + 2 α)}{s i n θ}$ and apply componendo and dividendo.]

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : m s i n θ = n s i n (θ + 2 α) \\ \Rightarrow \frac{s i n (θ + 2 α)}{s i n θ} = \frac{m}{n} \\ Using c o m p o n e n d o a n d d i v i d e n d o t h e o r m w e g e t \\ \Rightarrow \frac{s i n (θ + 2 α) + s i n θ}{s i n (θ + 2 α) - s i n θ} = \frac{m + n}{m - n} \\ \Rightarrow \frac{2 s i n (\frac{θ + 2 α + θ}{2}) . c o s (\frac{θ + 2 α - θ}{2})}{2 c o s (\frac{θ + 2 α + θ}{2}) . s i n (\frac{θ + 2 α - θ}{2})} = \frac{m + n}{m - n} \\ [\begin{array}{l} ? s i n A + s i n B = 2 s i n \frac{A + B}{2} . c o s \frac{A - B}{2} \\ s i n A - s i n B = 2 c o s \frac{A + B}{2} . s i n \frac{A - B}{2} \end{array}] \\ \Rightarrow \frac{s i n (θ + α) . c o s α}{c o s (θ + α) - s i n α} = \frac{m + n}{m - n} \\ \Rightarrow t a n (θ + α) . c o t α = \frac{m + n}{m - n} \\ H e n c e p r o v e d . \end{array}$

If $c o s (α + β) = \frac{4}{5}$ and $s i n (α - β) = \frac{5}{13}$ , where $α$ lies between $0$ and $\frac{π}{4}$ , find the value of $t a n^{2} α$ .
[Hint: Express $t a n 2 α$ as $t a n (α + β + α - β)$ .]

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ c o s (α + β) = \frac{4}{5} ∴ t a n (α + β) = \frac{3}{4} \\ a n d s i n (α - β) = \frac{5}{13} ∴ t a n (α - β) = \frac{5}{12} \\ N o w t a n 2 α = t a n [α + β + α - β] \\ = t a n [(α + β) + (α - β)] \\ = \frac{t a n (α + β) + t a n (α - β)}{1 - t a n (α + β) . t a n (α - β)} \\ = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \times \frac{5}{12}} = \frac{\frac{9 + 5}{12}}{\frac{48 - 15}{48}} = \frac{14}{12} \times \frac{48}{33} = \frac{56}{33} \\ H e n c e, t a n 2 α = \frac{56}{33} . \end{array}$

If $t a n x = \frac{b}{a}$ , then find the value of $\sqrt[]{\frac{a + b}{a - b}} + \sqrt[]{\frac{a - b}{a + b}} .$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n x = \frac{b}{a} \\ \sqrt[]{\frac{a + b}{a - b}} + \sqrt[]{\frac{a - b}{a + b}} = \frac{\sqrt[]{a + b}}{\sqrt[]{a - b}} + \frac{\sqrt[]{a - b}}{\sqrt[]{a + b}} \\ = \frac{a + b + a - b}{\sqrt[]{(a - b) (a + b)}} = \frac{2 a}{\sqrt[]{a^{2} - b^{2}}} = \frac{2 a}{a \sqrt[]{1 - \frac{b^{2}}{a^{2}}}} \\ = \frac{2}{\sqrt[]{1 - t a n^{2} x}} [? t a n x = \frac{b}{a}] \\ = \frac{2}{\sqrt[]{1 - \frac{s i n^{2} x}{c o s^{2} x}}} = \frac{2}{\frac{\sqrt[]{c o s^{2} x - s i n^{2} x}}{c o s x}} \\ = \frac{2 c o s x}{\sqrt[]{c o s 2 x}} [? c o s 2 x = c o s^{2} x - s i n^{2} x] \\ H e n c e, \sqrt[]{\frac{a + b}{a - b}} + \sqrt[]{\frac{a - b}{a + b}} = \frac{2 c o s x}{\sqrt[]{c o s 2 x}} \end{array}$

Prove that $c o s θ + c o s \frac{θ}{2} - c o s 3 θ c o s \frac{9 θ}{2} = s i n 7 θ s i n 8 θ .$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L . H . S . c o s θ c o s \frac{θ}{2} - c o s 3 θ c o s \frac{9 θ}{2} \\ = \frac{1}{2} [2 c o s θ c o s \frac{θ}{2}] - \frac{1}{2} [2 c o s 3 θ c o s \frac{9 θ}{2}] \\ = \frac{1}{2} [c o s (θ + \frac{θ}{2}) + c o s (θ - \frac{θ}{2})] - \frac{1}{2} [c o s (3 θ + \frac{9 θ}{2}) + c o s (3 θ - \frac{9 θ}{2})] \\ = \frac{1}{2} [c o s \frac{3 θ}{2} + c o s \frac{θ}{2} - c o s \frac{15 θ}{2} - c o s (\frac{- 3 θ}{2})] \\ = \frac{1}{2} [c o s \frac{3 θ}{2} + c o s \frac{θ}{2} - c o s \frac{15 θ}{2} - c o s \frac{3 θ}{2}] [? c o s (- θ) = c o s θ] \\ = \frac{1}{2} [c o s \frac{θ}{2} - c o s \frac{15 θ}{2}] = \frac{1}{2} [- 2 s i n (\frac{θ}{2} + \frac{15 θ}{2}) . s i n (\frac{θ}{2} - \frac{15 θ}{2})] \\ = - s i n 8 θ s i n (- 7 θ) = s i n 7 θ s i n 8 θ [? s i n (- θ) = - s i n θ] \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

If $a c o s θ + b s i n θ = m$ and $a s i n θ - b c o s θ = n$ , then show that $a^{2} + b^{2} = m^{2} + n^{2} .$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : a c o s θ + b s i n θ = m a n d a s i n θ - b c o s θ = n \\ R . H . S . m^{2} + n^{2} = {(a c o s θ + b s i n θ)}^{2} + {(a s i n θ - b c o s θ)}^{2} \\ = a^{2} c o s^{2} θ + b^{2} s i n^{2} θ + 2 a b s i n θ c o s θ + a^{2} s i n^{2} θ + b^{2} c o s^{2} θ - 2 a b s i n θ c o s θ \\ = a^{2} c o s^{2} θ + b^{2} s i n^{2} θ + a^{2} s i n^{2} θ + b^{2} c o s^{2} θ \\ = a^{2} (c o s^{2} θ + s i n^{2} θ) + b^{2} (s i n^{2} θ + c o s^{2} θ) \\ = a^{2} . 1 + b^{2} . 1 = a^{2} + b^{2} L . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

Find the value of $t a n 2 2^{\circ} 3 0^{'}$ .
[Hint: Use $θ = 4 5^{\circ}$ .]

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t 2 2^{0} 3 0^{'} = \frac{θ}{2} ∴ θ = 4 5^{0} \\ t a n 2 2^{0} 3 0^{'} = t a n \frac{θ}{2} = \frac{s i n \frac{θ}{2}}{c o s \frac{θ}{2}} = \frac{2 s i n \frac{θ}{2} c o s \frac{θ}{2}}{2 c o s^{2} \frac{θ}{2}} = \frac{s i n θ}{1 + c o s θ} \\ P u t θ = 4 5^{0} \\ ∴ \frac{s i n θ}{1 + c o s θ} = \frac{s i n 4 5^{0}}{1 + c o s 4 5^{0}} = \frac{\frac{1}{\sqrt[]{2}}}{1 + \frac{1}{\sqrt[]{2}}} = \frac{1}{\sqrt[]{2} + 1} \\ = \frac{1 \times (\sqrt[]{2} - 1)}{(\sqrt[]{2} + 1) (\sqrt[]{2} - 1)} = \sqrt[]{2} - 1 \\ H e n c e, t a n 2 2^{0} 3 0^{'} = \sqrt[]{2} - 1 . \end{array}$

Prove that $s i n 4 A = 4 s i n A c o s^{3} A - 4 c o s A s i n^{3} A .$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L . H . S . s i n 4 A = s i n (A + 3 A) \\ = s i n A c o s 3 A + c o s A s i n 3 A \\ = s i n A (4 c o s^{3} A - 3 c o s A) + c o s A (3 s i n A - 4 s i n^{3} A) \\ = 4 s i n A c o s^{3} A - 3 s i n A c o s A + 3 s i n A c o s A - 4 c o s A s i n^{3} A \\ = 4 s i n A c o s^{3} A - 4 c o s A s i n^{3} A R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

If $t a n θ + s i n θ = m$ and $t a n θ - s i n θ = n$ , then prove that $m^{2} - n^{2} = 4 s i n θ t a n θ .$

[Hint: Use $m + n = 2 t a n θ$ and $m - n = 2 s i n θ$ ., then use $m^{2} - n^{2} = (m + n) (m - n)$ ]

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ + s i n θ = m a n d t a n θ - s i n θ = n \\ L . H . S . m^{2} - n^{2} = (m + n) (m - n) \\ = [(t a n θ + s i n θ) + (t a n θ - s i n θ)] . [(t a n θ + s i n θ) - (t a n θ - s i n θ)] \\ = [t a n θ + s i n θ + t a n θ - s i n θ] . [t a n θ + s i n θ - t a n θ + s i n θ] \\ = 2 t a n θ . 2 s i n θ = 4 s i n θ t a n θ R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

If $t a n (A + B) = p$ , $t a n (A - B) = q$ , then show that $t a n 2 A = \frac{1 + p q}{p - q} .$

[Hint: Use $2 A = (A + B) + (A - B)$ .]

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n (A + B) = p a n d t a n (A - B) = q \\ L . H . S . t a n 2 A = t a n (A + B + A - B) = t a n [(A + B) + (A - B)] \\ = \frac{t a n (A + B) + t a n (A - B)}{1 - t a n (A + B) . t a n (A - B)} \\ = \frac{p + q}{1 - p q} R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

If $c o s α + c o s β = 0$ and $s i n α + s i n β = 0$ , then prove that $c o s 2 α + c o s 2 β = - 2 c o s (α + β) .$

[Hint: Use ${(c o s α + c o s β)}^{2} - {(s i n α + s i n β)}^{2} = 0$ .]

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s α + c o s β = 0 a n d s i n α + s i n β = 0 \\ ∴ {(c o s α + c o s β)}^{2} - {(s i n α + s i n β)}^{2} = 0 \\ \Rightarrow (c o s^{2} α + c o s^{2} β + 2 c o s α c o s β) - (s i n^{2} α + s i n^{2} β + 2 s i n α s i n β) = 0 \\ \Rightarrow c o s^{2} α + c o s^{2} β + 2 c o s α c o s β - s i n^{2} α - s i n^{2} β - 2 s i n α s i n β = 0 \\ \Rightarrow (c o s^{2} α - s i n^{2} α) + (c o s^{2} β - s i n^{2} β) + 2 (c o s α c o s β - s i n α s i n β) = 0 \\ \Rightarrow c o s 2 α + c o s 2 β + 2 c o s (α + β) = 0 \\ H e n c e, c o s 2 α + c o s 2 β = - 2 c o s (α + β) . \\ H e n c e p r o v e d . \end{array}$

If $\frac{s i n (x + y)}{a + b} = \frac{s i n (x - y)}{a - b},$ then show that $t a n x \cdot a = t a n y \cdot b .$

[Hint: Use Componendo and Dividendo.]

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \frac{s i n (x + y)}{s i n (x - y)} = \frac{a + b}{a - b} \\ \Rightarrow \frac{s i n (x + y) + s i n (x - y)}{s i n (x + y) - s i n (x - y)} = \frac{a + b + a - b}{a + b - a + b} [Using c o m p o n e n d o a n d d i v i d e n d o t h e o r e m] \\ \Rightarrow \frac{2 s i n (\frac{x + y + x - y}{2}) c o s (\frac{x + y - x + y}{2})}{2 c o s (\frac{x + y + x - y}{2}) s i n (\frac{x + y - x + y}{2})} = \frac{2 a}{2 b} \\ \Rightarrow \frac{s i n x . c o s y}{c o s x . s i n y} = \frac{a}{b} \Rightarrow t a n x . c o t y = \frac{a}{b} \\ \Rightarrow \frac{t a n x}{t a n y} = \frac{a}{b} . H e n c e p r o v e d . \end{array}$

If $t a n θ = \frac{s i n α - α}{c o s α + α},$ then show that $s i n α + c o s α = \sqrt[]{2} c o s θ .$

[Hint: Express the given equation as $t a n θ = t a n (α - \frac{π}{4})$ .]

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ = \frac{s i n α - c o s α}{s i n α + c o s α} \\ \Rightarrow t a n θ = \frac{t a n α - 1}{t a n α + 1} = \frac{t a n α - t a n \frac{π}{4}}{1 + t a n \frac{π}{4} t a n α} \\ \Rightarrow t a n θ = t a n (α - \frac{π}{4}) \\ ∴ θ = α - \frac{π}{4} \Rightarrow c o s θ = c o s (α - \frac{π}{4}) \\ \Rightarrow c o s θ = c o s α c o s \frac{π}{4} + s i n α s i n \frac{π}{4} \\ \Rightarrow c o s θ = c o s α . \frac{1}{\sqrt[]{2}} + s i n α . \frac{1}{\sqrt[]{2}} \\ \Rightarrow \sqrt[]{2} c o s θ = c o s α + s i n α \\ \Rightarrow c o s α + s i n α = \sqrt[]{2} c o s θ . H e n c e p r o v e d . \end{array}$

If $s i n θ + c o s θ = 1$ , then find the general value of $θ$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n θ + c o s θ = 1 \\ D i v i d i n g b o t h s i d e s b y \sqrt[]{{(1)}^{2} + {(1)}^{2}} = \sqrt[]{2} w e g e t \\ \frac{1}{\sqrt[]{2}} s i n θ + \frac{1}{\sqrt[]{2}} c o s θ = \frac{1}{\sqrt[]{2}} \dots (i) \\ \Rightarrow s i n \frac{π}{4} s i n θ + c o s \frac{π}{4} c o s θ = \frac{1}{\sqrt[]{2}} \\ \Rightarrow c o s (θ - \frac{π}{4}) = c o s \frac{π}{4} \\ \Rightarrow θ - \frac{π}{4} = 2 n π \pm \frac{π}{4}, n \in Z [\begin{array}{l} I f c o s θ = c o s α \\ θ = 2 n π \pm α \end{array}] \\ \Rightarrow θ = 2 n π + \frac{π}{4} + \frac{π}{4} o r θ = 2 n π - \frac{π}{4} + \frac{π}{4} \\ ∴ θ = 2 n π + \frac{π}{2} o r θ = 2 n π, n \in Z \\ H e n c e, t h e g e n e r a l v a l u e s o f θ a r e 2 n π + \frac{π}{2} a n d 2 n π . \end{array}$

Find the most general value of $θ$ satisfying the equation $t a n θ = - 1 and c o s θ = \frac{1}{\sqrt[]{2}} .$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ = - 1 a n d c o s θ = \frac{1}{\sqrt[]{2}} \\ t a n θ = - 1 \\ \Rightarrow t a n θ = t a n (\frac{- π}{4}) \\ \Rightarrow t a n θ = t a n (2 π - \frac{π}{4}) \Rightarrow t a n θ = t a n \frac{7 π}{4} \\ ∴ θ = \frac{7 π}{4} \\ N o w, c o s θ = \frac{1}{\sqrt[]{2}} \Rightarrow c o s θ = c o s \frac{π}{4} \\ \Rightarrow c o s θ = c o s (2 π - \frac{π}{4}) \\ \Rightarrow c o s θ = c o s \frac{7 π}{4} \\ ∴ θ = \frac{7 π}{4} [t a n θ a n d c o s θ a r e p o s i t i v e i n 4 t h q u a d r a n t] \\ H e n c e, t h e m o s t g e n e r a l v a l u e o f θ a r e 2 n π + \frac{7 π}{2} . \end{array}$

If $c o t θ + t a n θ = 2 c s c θ$ , then find the general value of $θ$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o t θ + t a n θ = 2 c o s e c θ \\ \Rightarrow \frac{c o s θ}{s i n θ} + \frac{s i n θ}{c o s θ} = \frac{2}{s i n θ} \\ \Rightarrow \frac{c o s^{2} θ + s i n^{2} θ}{s i n θ c o s θ} = \frac{2}{s i n θ} \\ \Rightarrow \frac{1}{s i n θ c o s θ} = \frac{2}{s i n θ} \\ \Rightarrow 2 s i n θ c o s θ = s i n θ \\ \Rightarrow 2 s i n θ c o s θ - s i n θ = 0 \\ \Rightarrow s i n θ (2 c o s θ - 1) = 0 \\ \Rightarrow s i n θ \neq 0 o r 2 c o s θ - 1 = 0 o r c o s θ = \frac{1}{2} \\ \Rightarrow c o s θ = c o s \frac{π}{3} \\ ∴ θ = 2 n π \pm \frac{π}{3} \\ H e n c e, t h e g e n e r a l v a l u e o f θ a r e 2 n π \pm \frac{π}{3} . \end{array}$

If $2 s i n^{2} θ = 3 c o s θ$ , where $0 \leq θ \leq 2 π$ , then find the value of $θ$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : 2 s i n^{2} θ = 3 c o s θ \\ \Rightarrow 2 (1 - c o s^{2} θ) = 3 c o s θ \\ \Rightarrow 2 - 2 c o s^{2} θ - 3 c o s θ = 0 \\ \Rightarrow 2 c o s^{2} θ + 3 c o s θ - 2 = 0 \\ \Rightarrow 2 c o s^{2} θ + 4 c o s θ - c o s θ - 2 = 0 \\ \Rightarrow 2 c o s θ (c o s θ + 2) - 1 (c o s θ + 2) = 0 \\ \Rightarrow (c o s θ + 2) (2 c o s θ - 1) = 0 \\ \Rightarrow c o s θ + 2 = 0 o r 2 c o s θ - 1 = 0 \\ \Rightarrow c o s θ \neq - 2 [- 1 \leq c o s θ \leq 1] \\ ∴ 2 c o s θ - 1 = 0 \\ \Rightarrow c o s θ = \frac{1}{2} \\ \Rightarrow c o s θ = c o s \frac{π}{3}, c o s (2 π - \frac{π}{3}) \\ \Rightarrow c o s θ = c o s \frac{π}{3}, c o s \frac{5 π}{3} \\ ∴ θ = 2 n π \pm \frac{π}{3} a n d θ = 2 n π \pm \frac{5 π}{3} (n \in Z) \\ H e n c e, t h e v a l u e o f θ a r e \frac{π}{3} a n d \frac{5 π}{3} . \end{array}$

If $s e c x c o s 5 x + 1 = 0,$ where $0 < x \leq \frac{π}{2}$ , then find the value of $x$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s e c x c o s 5 x + 1 = 0 \\ \Rightarrow \frac{1}{c o s x} . c o s 5 x + 1 = 0 \\ \Rightarrow c o s 5 x + c o s x = 0 \\ \Rightarrow 2 c o s (\frac{5 x + x}{2}) . c o s (\frac{5 x - x}{2}) = 0 \\ \Rightarrow c o s 3 x . c o s 2 x = 0 \\ \Rightarrow c o s 3 x = 0 o r c o s 2 x = 0 \\ \Rightarrow 3 x = \frac{π}{2} o r 2 x = \frac{π}{2} \\ \Rightarrow x = \frac{π}{6} o r x = \frac{π}{4} \\ H e n c e, t h e v a l u e o f x a r e \frac{π}{6}, \frac{π}{4} . \end{array}$

Trigonometric Functions Long Answer Type Questions

1. If $s i n (θ + α) = a$ and $s i n (θ + β) = b$ , then prove that $c o s 2 (α - β) - 4 a b c o s (α - β) = 1 - 2 a^{2} - 2 b^{2} .$

Hint: Express cos $(α - β) =$ cos (( $θ +$ $α$ ) $-$ $(θ + β)$ )]

Sol:

$\begin{array}{l} G i v e n t h a t : \\ s i n (θ + α) = a a n d s i n (θ + β) = b \\ c o s (α - β) = c o s [θ + α - θ - β] = c o s [(θ + α) - (θ + β)] \\ = c o s (θ + α) c o s (θ + β) + s i n (θ + α) s i n (θ + β) \\ = \sqrt[]{1 - s i n^{2} (θ + α)} \sqrt[]{1 - s i n^{2} (θ + β)} + a . b \\ = \sqrt[]{(1 - a^{2}) (1 - b^{2})} + a b = a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ N o w c o s 2 (α - β) - 4 a b c o s (α - β) \\ = 2 c o s^{2} (α - β) - 1 - 4 a b c o s (α - β) [∵ c o s 2 θ = 2 c o s^{2} θ - 1] \\ = 2 {[a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}]}^{2} - 1 - 4 a b [a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}] \\ = 2 [a^{2} b^{2} + 1 - a^{2} - b^{2} + a^{2} b^{2} + 2 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}] - 1 - 4 a^{2} b^{2} - 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ = 2 a^{2} b^{2} + 2 - 2 a^{2} - 2 b^{2} + 2 a^{2} b^{2} + 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} - 1 - 4 a^{2} b^{2} - 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ = 1 - 2 a^{2} - 2 b^{2} \\ H e n c e, c o s 2 (α - β) - 4 a b c o s (α - β) = 1 - 2 a^{2} - 2 b^{2} \\ H e n c e p r o v e d . \end{array}$

2. If $c o s (θ + ϕ) = m c o s (θ - ϕ)$ , then prove that tan $θ = \frac{1 - m}{1 + m} . c o t ϕ$

[Hint: Express $\frac{c o s (θ + α)}{c o s (θ - α)} = \frac{m}{1}$ .]and apply componendo and dividendo.]

Sol:

$\begin{array}{l} G i v e n t h a t : c o s (θ + ϕ) = m c o s (θ - ϕ) \\ \Rightarrow \frac{c o s (θ + ϕ)}{c o s (θ - ϕ)} = \frac{m}{1} \\ Using c o m p o n e n d o a n d d i v i d e n d o t h e o r e m, w e g e t \\ \frac{c o s (θ + ϕ) + c o s (θ - ϕ)}{c o s (θ + ϕ) - c o s (θ - ϕ)} = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{2 c o s (\frac{θ + ϕ + θ - ϕ}{2}) . c o s (\frac{θ + ϕ - θ + ϕ}{2})}{- 2 s i n (\frac{θ + ϕ + θ - ϕ}{2}) . s i n (\frac{θ + ϕ - θ + ϕ}{2})} = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{c o s θ . c o s ϕ}{- s i n θ . s i n ϕ} = \frac{m + 1}{m - 1} \Rightarrow - c o t θ . c o t ϕ = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{- c o t ϕ}{t a n θ} = \frac{m + 1}{m - 1} = - \frac{1 + m}{1 - m} \\ \Rightarrow t a n θ = \frac{1 + m}{1 - m} c o t ϕ . \\ H e n c e p r o v e d . \end{array}$

Commonly asked questions

If $s i n (θ + α) = a$ and $s i n (θ + β) = b$ , then prove that $c o s 2 (α - β) - 4 a b c o s (α - β) = 1 - 2 a^{2} - 2 b^{2} .$

Hint: Express cos $(α - β) =$ cos (( $θ +$ $α$ ) $-$ $(θ + β)$ )]

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ s i n (θ + α) = a a n d s i n (θ + β) = b \\ c o s (α - β) = c o s [θ + α - θ - β] = c o s [(θ + α) - (θ + β)] \\ = c o s (θ + α) c o s (θ + β) + s i n (θ + α) s i n (θ + β) \\ = \sqrt[]{1 - s i n^{2} (θ + α)} \sqrt[]{1 - s i n^{2} (θ + β)} + a . b \\ = \sqrt[]{(1 - a^{2}) (1 - b^{2})} + a b = a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ N o w c o s 2 (α - β) - 4 a b c o s (α - β) \\ = 2 c o s^{2} (α - β) - 1 - 4 a b c o s (α - β) [? c o s 2 θ = 2 c o s^{2} θ - 1] \\ = 2 {[a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}]}^{2} - 1 - 4 a b [a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}] \\ = 2 [a^{2} b^{2} + 1 - a^{2} - b^{2} + a^{2} b^{2} + 2 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}] - 1 - 4 a^{2} b^{2} - 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ = 2 a^{2} b^{2} + 2 - 2 a^{2} - 2 b^{2} + 2 a^{2} b^{2} + 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} - 1 - 4 a^{2} b^{2} - 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ = 1 - 2 a^{2} - 2 b^{2} \\ H e n c e, c o s 2 (α - β) - 4 a b c o s (α - β) = 1 - 2 a^{2} - 2 b^{2} \\ H e n c e p r o v e d . \end{array}$

If $c o s (θ + ϕ) = m c o s (θ - ϕ)$ , then prove that tan $θ = \frac{1 - m}{1 + m} . c o t ϕ$

[Hint: Express $\frac{c o s (θ + α)}{c o s (θ - α)} = \frac{m}{1}$ .]and apply componendo and dividendo.]

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s (θ + ?) = m c o s (θ - ?) \\ \Rightarrow \frac{c o s (θ + ?)}{c o s (θ - ?)} = \frac{m}{1} \\ Using c o m p o n e n d o a n d d i v i d e n d o t h e o r e m, w e g e t \\ \frac{c o s (θ + ?) + c o s (θ - ?)}{c o s (θ + ?) - c o s (θ - ?)} = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{2 c o s (\frac{θ + ? + θ - ?}{2}) . c o s (\frac{θ + ? - θ + ?}{2})}{- 2 s i n (\frac{θ + ? + θ - ?}{2}) . s i n (\frac{θ + ? - θ + ?}{2})} = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{c o s θ . c o s ?}{- s i n θ . s i n ?} = \frac{m + 1}{m - 1} \Rightarrow - c o t θ . c o t ? = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{- c o t ?}{t a n θ} = \frac{m + 1}{m - 1} = - \frac{1 + m}{1 - m} \\ \Rightarrow t a n θ = \frac{1 + m}{1 - m} c o t ? . \\ H e n c e p r o v e d . \end{array}$

Find the value of the expression $3 [s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)] - 2 [s i n^{6} (2 π / 2 + α) + s i n^{6} (5 π - α)] .$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ 3 [s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)] - 2 [s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)] \\ = 3 [c o s^{4} α + s i n^{4} (π + α)] - 2 [c o s^{6} α + s i n^{6} (π - α)] \\ = 3 [c o s^{4} α + s i n^{4} α] - 2 [c o s^{6} α + s i n^{6} α] \\ = 3 [c o s^{4} α + s i n^{4} α + 2 s i n^{2} α c o s^{2} α - 2 s i n^{2} α c o s^{2} α] - 2 [\begin{array}{l} {(c o s^{2} α + s i n^{2} α)}^{3} \\ - 3 c o s^{2} α s i n^{2} α (c o s^{2} α + s i n^{2} α) \end{array}] \\ = 3 [{(c o s^{2} α + s i n^{2} α)}^{2} - 2 s i n^{2} α c o s^{2} α] - 2 [1 - 3 c o s^{2} α s i n^{2} α] \\ = 3 [1 - 2 s i n^{2} α c o s^{2} α] - 2 [1 - 3 c o s^{2} α s i n^{2} α] \\ = 3 - 6 s i n^{2} α c o s^{2} α - 2 + 6 c o s^{2} α s i n^{2} α \\ = 3 - 2 = 1 \\ H e n c e, t h e v a l u e o f t h e g i v e n expression i s 1 . \end{array}$

If $a c o s 2 θ + b s i n 2 θ = c,$ has $α$ and $β$ as its roots, then prove that $t a n α + t a n β = \frac{2 b}{a + c} .$

[Hint: Use the identities for $c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}$ and $s i n 2 θ$ = $\frac{2 t a n θ}{1 + t a n^{2} θ}$ ]

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : a c o s 2 θ + b s i n 2 θ = c \\ \Rightarrow a [\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}] + b [\frac{2 t a n θ}{1 + t a n^{2} θ}] = c \\ \Rightarrow a - a t a n^{2} θ + 2 b t a n θ = c (1 + t a n^{2} θ) [? c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}, s i n 2 θ = \frac{2 t a n θ}{1 + t a n^{2} θ}] \\ \Rightarrow a - a t a n^{2} θ + 2 b t a n θ - c t a n^{2} θ - c = 0 \\ \Rightarrow - (a + c) t a n^{2} θ + 2 b t a n θ + (a - c) = 0 \\ \Rightarrow (a + c) t a n^{2} θ - 2 b t a n θ + (c - a) = 0 \\ Since α a n d β a r e t h e r o o t s o f t h i s e q u a t i o n \\ \Rightarrow t a n α + t a n β = \frac{- (- 2 b)}{a + c} \\ \Rightarrow t a n α + t a n β = \frac{2 b}{a + c} . H e n c e p r o v e d . \end{array}$

If $θ$ lies in the first quadrant and $c o s θ = \frac{8}{17}$ , then find the value of $c o s (3 0^{\circ} + θ) + c o s (4 5^{\circ} - θ) + c o s (12 0^{\circ} - θ) .$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s θ = \frac{8}{17} \\ ∴ s i n θ = \sqrt[]{1 - {(\frac{8}{17})}^{2}} = \sqrt[]{1 - \frac{64}{289}} = \sqrt[]{\frac{289 - 64}{289}} = \sqrt[]{\frac{225}{289}} = \frac{15}{17} \\ B u t θ l i e s i n I q u a d r a n t . \\ ∴ s i n θ = \frac{15}{17} \\ N o w c o s (3 0^{0} + θ) + c o s (4 5^{0} - θ) + c o s (12 0^{0} - θ) \\ = c o s 3 0^{0} c o s θ - s i n 3 0^{0} s i n θ + c o s 4 5^{0} c o s θ + s i n 4 5^{0} s i n θ + c o s 12 0^{0} c o s θ + s i n 12 0^{0} s i n θ \\ = \frac{\sqrt[]{3}}{2} c o s θ - \frac{1}{2} s i n θ + \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ - \frac{1}{2} c o s θ + \frac{\sqrt[]{3}}{2} s i n θ \\ = (\frac{\sqrt[]{3}}{2} c o s θ + \frac{\sqrt[]{3}}{2} s i n θ) - \frac{1}{2} (s i n θ + c o s θ) + \frac{1}{\sqrt[]{2}} (c o s θ + s i n θ) \\ = \frac{\sqrt[]{3}}{2} (c o s θ + s i n θ) - \frac{1}{2} (s i n θ + c o s θ) + \frac{1}{\sqrt[]{2}} (c o s θ + s i n θ) \\ = (\frac{\sqrt[]{3}}{2} - \frac{1}{2} + \frac{1}{\sqrt[]{2}}) (c o s θ + s i n θ) = (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) (\frac{8}{17} + \frac{15}{17}) \\ = (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) (\frac{23}{17}) \\ = \frac{23}{17} (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) . \\ H e n c e, t h e r e q u i r e d s o l u t i o n = \frac{23}{17} (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) . \end{array}$

Find the value of the expression $c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) + c o s^{4} (\frac{5 π}{8}) + c o s^{4} (\frac{7 π}{8}) .$

Hint : Simplify the expression to 2 ( $c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) = 2 [{(c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8})}^{2} - c o s^{2} \frac{π}{8} c o s^{2} \frac{3 π}{8} .$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) + c o s^{4} (\frac{5 π}{8}) + c o s^{4} (\frac{7 π}{8}) . \\ = c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) + c o s^{4} (π - \frac{3 π}{8}) + c o s^{4} (π - \frac{π}{8}) \\ = c o s^{4} \frac{π}{8} + c o s^{4} \frac{3 π}{8} + c o s^{4} \frac{3 π}{8} + c o s^{4} \frac{π}{8} \\ = 2 c o s^{4} \frac{π}{8} + 2 c o s^{4} \frac{3 π}{8} = 2 [c o s^{4} \frac{π}{8} + c o s^{4} \frac{3 π}{8}] \\ = 2 [c o s^{4} \frac{π}{8} + c o s^{4} (\frac{π}{2} - \frac{π}{8})] = 2 [c o s^{4} \frac{π}{8} + s i n^{4} \frac{π}{8}] \\ = 2 [c o s^{4} \frac{π}{8} + s i n^{4} \frac{π}{8} + 2 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8} - 2 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8}] \\ = 2 [{(c o s^{2} \frac{π}{8} + s i n^{2} \frac{π}{8})}^{2} - 2 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8}] \\ = 2 [1 - 2 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8}] = 2 - 4 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8} \\ = 2 - {(2 s i n \frac{π}{8} . c o s \frac{π}{8})}^{2} = 2 - {(s i n \frac{π}{4})}^{2} \\ = 2 - {(\frac{1}{\sqrt[]{2}})}^{2} = 2 - \frac{1}{2} = \frac{3}{2} \\ H e n c e, t h e r e q u i r e d v a l u e o f t h e expression = \frac{3}{2} . \end{array}$

Find the general solution of the equation $5 c o s^{2} θ + 7 s i n^{2} θ - 6 = 0 .$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} 5 c o s^{2} θ + 7 s i n^{2} θ - 6 = 0 \\ \Rightarrow 5 c o s^{2} θ + 7 (1 - c o s^{2} θ) - 6 = 0 \\ \Rightarrow 5 c o s^{2} θ + 7 - 7 c o s^{2} θ - 6 = 0 \Rightarrow - 2 c o s^{2} θ + 1 = 0 \\ \Rightarrow 2 c o s^{2} θ + 1 = 0 \Rightarrow c o s^{2} θ = \frac{1}{2} \\ \Rightarrow c o s^{2} θ = c o s^{2} \frac{π}{4} \\ ∴ θ = n π \pm \frac{π}{4} [\begin{array}{l} ? I f c o s^{2} θ = c o s^{2} α \\ ∴ θ = n π \pm α \end{array}] \\ H e n c e, t h e g e n e r a l s o l u t i o n o f θ = n π \pm \frac{π}{4}, n \in Z . \end{array}$

Find the general solution of the equation $s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x .$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x \\ \Rightarrow (s i n 3 x + s i n x) - 3 s i n 2 x = (c o s 3 x + c o s x) - 3 c o s 2 x \\ \Rightarrow 2 s i n (\frac{3 x + x}{2}) . c o s (\frac{3 x - x}{2}) - 3 s i n 2 x = 2 c o s (\frac{3 x + x}{2}) . c o s (\frac{3 x - x}{2}) - 3 c o s 2 x \\ \Rightarrow 2 s i n 2 x . c o s x - 3 s i n 2 x = 2 c o s 2 x . c o s x - 3 c o s 2 x \\ \Rightarrow 2 s i n 2 x . c o s x - 2 c o s 2 x . c o s x = 3 s i n 2 x - 3 c o s 2 x \\ \Rightarrow 2 c o s x (s i n 2 x - c o s 2 x) - 3 (s i n 2 x - c o s 2 x) = 0 \\ \Rightarrow (s i n 2 x - c o s 2 x) (2 c o s x - 3) = 0 \\ \Rightarrow s i n 2 x - c o s 2 x = 0 a n d 2 c o s x - 3 \neq 0 [? - 1 \leq c o s x \leq 1] \\ \Rightarrow \frac{s i n 2 x}{c o s 2 x} - 1 = 0 \Rightarrow t a n 2 x = 1 \\ \Rightarrow t a n 2 x = t a n \frac{π}{4} \Rightarrow 2 x = n π + \frac{π}{4} ∴ x = \frac{n π}{2} + \frac{π}{8} \\ H e n c e, t h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n i s \\ x = \frac{n π}{2} + \frac{π}{8}, n \in Z . \end{array}$

Find the general solution of the equation

$(\sqrt[]{3} - 1) c o s θ + (\sqrt[]{3} + 1) s i n θ = 2 .$

[Hint: Put $\sqrt[]{3} - 1 = r s i n α, \sqrt[]{3} + 1 = r c o s α w h i c h g i v e s$ . ] $t a n α = t a n (\frac{π}{4} - \frac{π}{6}) α = \frac{π}{12}]$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : (\sqrt[]{3} - 1) c o s θ + (\sqrt[]{3} + 1) s i n θ = 2 \\ P u t \sqrt[]{3} - 1 = r s i n α, \sqrt[]{3} + 1 = r c o s α \\ S q u a r i n g a n d a d d i n g, w e g e t \\ r^{2} = 3 + 1 - 2 \sqrt[]{3} + 3 + 1 + 2 \sqrt[]{3} \\ r^{2} = 8 \Rightarrow r = \pm 2 \sqrt[]{2} \\ N o w t h e g i v e n e q u a t i o n c a n b e w r i t t e n a s \\ r s i n α c o s θ + r c o s α s i n θ = 2 \\ \Rightarrow r (s i n α c o s θ + c o s α s i n θ) = 2 \\ \Rightarrow 2 \sqrt[]{2} s i n (α + θ) = 2 \\ \Rightarrow s i n (α + θ) = \frac{2}{2 \sqrt[]{2}} = \frac{1}{\sqrt[]{2}} \\ \Rightarrow s i n (α + θ) = s i n \frac{π}{4} \\ \Rightarrow α + θ = n π + {(- 1)}^{n} . \frac{π}{4} \dots (i) \\ N o w \frac{r s i n α}{r c o s α} = \frac{\sqrt[]{3} - 1}{\sqrt[]{3} + 1} \\ \Rightarrow t a n α = \frac{t a n \frac{π}{3} - t a n \frac{π}{4}}{1 + t a n \frac{π}{4} . t a n \frac{π}{3}} \\ \Rightarrow t a n α = t a n (\frac{π}{3} - \frac{π}{4}) \\ \Rightarrow t a n α = t a n \frac{π}{12} ∴ α = \frac{π}{12} \\ P u t t i n g t h e v a l u e o f α i n e q n . (i) w e g e t \\ \frac{π}{12} + θ = n π + {(- 1)}^{n} . \frac{π}{4} \\ ∴ θ = n π + {(- 1)}^{n} . \frac{π}{4} - \frac{π}{12} \\ H e n c e, t h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n i s \\ θ = n π + {(- 1)}^{n} . \frac{π}{4} - \frac{π}{12}, n \in Z . \end{array}$

Trigonometric Functions Objective Type Questions

Choose the correct answer from the given four options in each of the Exercises 30 to 59:

1. If $s i n θ + c o s e c θ = 2$ , then $s i n^{2} θ + c o s e c^{2} θ$ is equal to:

(a) 1

(b) 4

(c) 2

(d) None of these

Sol:

$\begin{array}{l} G i v e n t h a t : s i n θ + c o s e c θ = 2 \\ S q u a r i n g b o t h s i d e s, w e g e t \\ {(s i n θ + c o s e c θ)}^{2} = {(2)}^{2} \\ \Rightarrow s i n^{2} θ + c o s e c^{2} θ + 2 s i n θ c o s e c θ = 4 \\ \Rightarrow s i n^{2} θ + c o s e c^{2} θ + 2 s i n θ \times \frac{1}{s i n θ} = 4 \\ \Rightarrow s i n^{2} θ + c o s e c^{2} θ + 2 = 4 \\ \Rightarrow s i n^{2} θ + c o s e c^{2} θ = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

2. If $f (x) = c o s^{2} x + s e c^{2} x$ , then:

(a) $f (x) < 1$

(b) $f (x) = 1$

(c) $2 < f (x) < 1$

(d) f(x) $\geq 2$

[Hint: A.M ≥ G.M.]

Sol:

$\begin{array}{l} G i v e n t h a t : f (x) = c o s^{2} x + se c^{2} x \\ W e k n o w t h a t A M \geq G M \\ \Rightarrow \frac{c o s^{2} x + se c^{2} x}{2} \geq \sqrt[]{c o s^{2} x . se c^{2} x} \\ \Rightarrow \frac{c o s^{2} x + se c^{2} x}{2} \geq 1 \Rightarrow c o s^{2} x + se c^{2} x \geq 2 \\ \Rightarrow f (x) \geq 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Commonly asked questions

Choose the correct answer from the given four options in each of the Exercises 30 to 59:

If $s i n θ + c o s e c θ = 2$ , then $s i n^{2} θ + c o s e c^{2} θ$ is equal to:

(a) 1

(b) 4

(c) 2

(d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

If $f (x) = c o s^{2} x + s e c^{2} x$ , then:

(a) $f (x) < 1$

(b) $f (x) = 1$

(c) $2 < f (x) < 1$

(d) f(x) $\geq 2$

[Hint: A.M ≥ G.M.]

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

If $t a n θ = \frac{1}{2} =$ and $t a n ϕ = 1 / 3$ , then the value of $θ + ϕ$ is:

(a) $\frac{π}{6}$

(b) $π$

(c) 0

(d) $\frac{π}{4}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t \\ t a n (θ + ?) = \frac{t a n θ + t a n ?}{1 - t a n θ t a n ?} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \\ \Rightarrow t a n (θ + ?) = t a n \frac{π}{4} \\ ∴ θ + ? = \frac{π}{4} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Which of the following is not correct?

(a) $s i n θ = - \frac{1}{5}$

(b) $c o s θ = 1$

(c) $s e c θ = 2$

(d) $t a n θ = 20$

This is an Objective Type Questions as classified in NCERT Exempar

Sol:

$\begin{array}{l} s i n θ = - \frac{1}{5} i s c o r r e c t . ? - 1 \leq s i n θ \leq 1 \\ S o (a) i s c o r r e c t . \\ c o s θ = 1 i s c o r r e c t . ? c o s 0^{0} = 1 \\ S o (b) i s c o r r e c t . \\ s e c θ = \frac{1}{2} \Rightarrow c o s θ = 2 i s n o t c o r r e c t . ? - 1 \leq c o s θ \leq 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The value of $t a n 1^{\circ} \cdot t a n 2^{\circ} \cdot t a n 3^{\circ} \dots t a n 8 9^{\circ}$ is:

(a) 0

(b) 1

(c) 2

(d) Not defined

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n 1^{0} t a n 2^{0} t a n 3^{0} \dots t a n 8 9^{0} \\ = t a n 1^{0} t a n 2^{0} t a n 3^{0} \dots t a n 4 5^{0} . t a n {(90 - 44)}^{0} . t a n {(90 - 43)}^{0} \dots t a n {(90 - 1)}^{0} \\ = t a n 1^{0} c o t 1^{0} . t a n 2^{0} c o t 2^{0} . t a n 3^{0} c o t 3^{0} \dots t a n 8 9^{0} . c o t 8 9^{0} \\ = 1.1.1.1 \dots 1.1 = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The value of $\frac{1 - t a n^{2} 1 5^{\circ}}{1 + t a n^{2} 15 \circ}$ is:

(a) 1

(b) $\sqrt[]{3}$

(c) $\frac{\sqrt[]{3}}{2}$

(d) 2

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \frac{1 - t a n^{2} 1 5^{0}}{1 + t a n^{2} 1 5^{0}} \\ L e t θ = 1 5^{0} ∴ 2 θ = 3 0^{0} \\ c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ} \\ \Rightarrow c o s 3 0^{0} = \frac{1 - t a n^{2} 1 5^{0}}{1 + t a n^{2} 1 5^{0}} \\ \Rightarrow \frac{\sqrt[]{3}}{2} = \frac{1 - t a n^{2} 1 5^{0}}{1 + t a n^{2} 1 5^{0}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The value of $c o s 1^{\circ} \cdot c o s 2^{\circ} \cdot c o s 3^{\circ} \dots c o s 17 9^{\circ}$ is:

(a) $\frac{1}{\sqrt[]{2}}$

(b) 0

(c) 1

(d) -1

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s 1^{0} . c o s 2^{0} . c o s 3^{0} \dots c o s 17 9^{0} \\ = c o s 1^{0} . c o s 2^{0} . c o s 3^{0} \dots c o s 9 0^{0} . c o s 9 1^{0} \dots c o s 17 9^{0} \\ = 0 [? c o s 9 0^{0} = 0] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If $t a n θ = 3$ and $θ$ lies in the third quadrant, then the value of $s i n θ$ is:

(a) $\frac{1}{\sqrt[]{10}}$

(b) $- \frac{1}{\sqrt[]{10}}$

(c) $\frac{- 3}{\sqrt[]{10}}$

(d) $\frac{3}{\sqrt[]{10}}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

\begin{array}{l} t a n θ = 3, θ l i e s i n t h i r d q u a d r a n t \\ ∴ s i n θ = \frac{- 3}{\sqrt[]{10}} w h e r e θ l i e s i n t h i r d q u a d r a n t \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}

The value of $t a n 7 5^{\circ} - c o t 7 5^{\circ}$ is equal to:

(a) $2 \sqrt[]{3}$

(b) $2 + \sqrt[]{3}$

(c) $2 - \sqrt[]{3}$

(d) 1

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression is t a n 7 5^{0} - c o t 7 5^{0} \\ t a n 7 5^{0} - c o t 7 5^{0} = t a n 7 5^{0} - c o t {(90 - 15)}^{0} \\ = t a n 7 5^{0} - t a n 1 5^{0} = \frac{s i n 7 5^{0}}{c o s 7 5^{0}} - \frac{s i n 1 5^{0}}{c o s 1 5^{0}} \\ = \frac{s i n 7 5^{0} c o s 1 5^{0} - c o s 7 5^{0} s i n 1 5^{0}}{c o s 7 5^{0} c o s 1 5^{0}} \\ = \frac{s i n (7 5^{0} - 1 5^{0})}{\frac{1}{2} \times 2 c o s 7 5^{0} c o s 1 5^{0}} \\ = \frac{2 s i n 6 0^{0}}{c o s (7 5^{0} + 1 5^{0}) + c o s (7 5^{0} - 1 5^{0})} \\ = \frac{2 \times \frac{\sqrt[]{3}}{2}}{c o s 9 0^{0} c o s 6 0^{0}} = \frac{\sqrt[]{3}}{0 + \frac{1}{2}} = 2 \sqrt[]{3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Which of the following is correct?

(a) $s i n 1^{\circ} > s i n 1$

(b) $s i n 1^{\circ} < s i n 1$

(c) $s i n 1^{\circ} = s i n 1$

(d) $s i n 1^{\circ} = \frac{π}{1 8^{0}}$

Hint 1 radian = $\frac{π}{18 0^{0}} = 5 7^{0} 3 0^{'} a p p r o x$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t i f θ i n c r e a s e s t h e n t h e v a l u e o f s i n θ a l s o i n c r e a s e s . \\ S o, s i n 1^{0} < s i n 1 [? 1 r a d i a n = \frac{π}{180} s i n 1] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If $t a n α = \frac{m}{m + 1}$ and $t a n β = \frac{1}{m + 2}$ , then $α + β$ is equal to:

(a) $\frac{π}{2}$

(b) $\frac{π}{3}$

(c) $\frac{π}{6}$

(d) $\frac{π}{4}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t t a n α = \frac{m}{m + 1} a n d t a n β = \frac{1}{2 m + 1} \\ t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α . t a n β} = \frac{\frac{m}{m + 1} + \frac{1}{2 m + 1}}{1 - \frac{m}{m + 1} \times \frac{1}{2 m + 1}} \\ = \frac{\frac{2 m^{2} + m + m + 1}{(m + 1) (2 m + 1)}}{\frac{(m + 1) (2 m + 1) - m}{(m + 1) (2 m + 1)}} = \frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + m + 1 - m} \\ = \frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + 1} = 1 \\ t a n (α + β) = t a n \frac{π}{4} ∴ α + β = \frac{π}{4} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The minimum value of $3 c o s x + 4 s i n x + 8$ is:

(a) 5

(b) 9

(c) 7

(d) 3

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s 3 c o s x + 4 s i n x + 8 \\ L e t y = 3 c o s x + 4 s i n x + 8 \\ \Rightarrow y - 8 = 3 c o s x + 4 s i n x \\ M i n i m u m v a l u e o f y - 8 = - \sqrt[]{{(3)}^{2} + {(4)}^{2}} \\ y - 8 = - \sqrt[]{9 + 16} = - 5 \\ y = 8 - 5 = 3 \\ S o, t h e m i n i m u m v a l u e o f t h e g i v e n expression i s 3 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The value of tan 3A -tan2A-tanA is equal to

(a) tan3Atan2AtanA

(b) -tan3Atan2AtanA

(c) tanAtan2A-tan2Atan3A-tan3AtanA

(d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s t a n 3 A - t a n 2 A - t a n A \\ t a n 3 A = t a n (2 A + A) \\ \Rightarrow t a n 3 A = \frac{t a n 2 A + t a n A}{1 - t a n 2 A . t a n A} \\ \Rightarrow t a n 3 A (1 - t a n 2 A . t a n A) = t a n 2 A + t a n A \\ \Rightarrow t a n 3 A - t a n 3 A t a n 2 A t a n A = t a n 2 A + t a n A \\ \Rightarrow t a n 3 A - t a n 2 A - t a n A = t a n 3 A t a n 2 A t a n A \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The value of $s i n (4 5^{\circ} + θ) - c o s (4 5^{\circ} - θ)$ is:

(a) $2 c o s θ$

(b) $2 s i n θ$

(c) 1

(d) 0

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s s i n (4 5^{0} + θ) - c o s (4 5^{0} - θ) \\ s i n (4 5^{0} + θ) = s i n 4 5^{0} c o s θ + c o s 4 5^{0} s i n θ \\ = \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ \\ c o s (4 5^{0} - θ) = c o s 4 5^{0} c o s θ + s i n 4 5^{0} s i n θ \\ = \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ \\ s i n (4 5^{0} + θ) - c o s (4 5^{0} - θ) = \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ - \frac{1}{\sqrt[]{2}} c o s θ - \frac{1}{\sqrt[]{2}} s i n θ = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The value of is: $c o t (\frac{π}{4} + θ) - c o t (\frac{π}{4} - θ)$

(a) $- 1$

(b) $0$

(c) $1$

(d) Not defined

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s c o t (\frac{π}{4} + θ) . c o t (\frac{π}{4} - θ) \\ = \frac{c o t \frac{π}{4} c o t θ - 1}{c o t θ + c o t \frac{π}{4}} \times \frac{c o t \frac{π}{4} c o t θ + 1}{c o t θ - c o t \frac{π}{4}} \\ = \frac{1 . c o t θ - 1}{c o t θ + 1} \times \frac{1 . c o t θ + 1}{c o t θ - 1} \\ = \frac{c o t θ - 1}{c o t θ + 1} \times \frac{c o t θ + 1}{c o t θ - 1} = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

$c o s 2 θ c o s 2 ϕ + s i n^{2} (θ - ϕ) - s i n^{2} (θ + ϕ)$ is equal to:

(a) $s i n 2 (θ + ϕ)$

(b) $c o s 2 (θ + ϕ)$

(c) $s i n 2 (θ - ϕ)$

(d) $c o s 2 (θ - ϕ)$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s 2 θ c o s 2 ? + s i n^{2} (θ - ?) - s i n^{2} (θ + ?) \\ c o s 2 θ c o s 2 ? + s i n^{2} (θ - ?) - s i n^{2} (θ + ?) = c o s 2 θ c o s 2 ? + s i n (θ - ? + θ + ?) . s i n (θ - ? - θ - ?) \\ [? s i n^{2} A - s i n^{2} B = s i n (A + B) . s i n (A - B)] \\ = c o s 2 θ c o s 2 ? + s i n 2 θ . s i n (- 2 ?) \\ = c o s 2 θ c o s 2 ? - s i n 2 θ . s i n 2 ? [? s i n (- θ) = - s i n θ] \\ = c o s (2 θ + 2 ?) = c o s 2 (θ + ?) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The value of $c o s 1 2^{?} + c o s 8 4^{?} + c o s 15 6^{?} + c o s 13 2^{?}$ is:

(a) $- \frac{1}{2}$

(b) $1$

(c) $\frac{1}{8}$

(d) $\frac{1}{2}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s 1 2^{0} + c o s 8 4^{0} + c o s 15 6^{0} + c o s 13 2^{0} \\ = (c o s 13 2^{0} + c o s 1 2^{0}) + (+ c o s 15 6^{0} + c o s 8 4^{0}) \\ = (2 c o s \frac{13 2^{0} + 1 2^{0}}{2} . c o s \frac{13 2^{0} - 1 2^{0}}{2}) + (2 c o s \frac{15 6^{0} + 8 4^{0}}{2} . c o s \frac{15 6^{0} - 8 4^{0}}{2}) \\ = 2 c o s 7 2^{0} . c o s 6 0^{0} + 2 c o s 12 0^{0} . c o s 3 6^{0} \\ = 2 c o s 7 2^{0} \times \frac{1}{2} + 2 \times (- \frac{1}{2}) c o s 3 6^{0} = c o s 7 2^{0} - c o s 3 6^{0} \\ = c o s (9 0^{0} - 1 8^{0}) - c o s 3 6^{0} = s i n 1 8^{0} - c o s 3 6^{0} \\ = \frac{\sqrt[]{5} - 1}{4} - \frac{\sqrt[]{5} + 1}{4} [? s i n 1 8^{0} = \frac{\sqrt[]{5} - 1}{4}, c o s 3 6^{0} = \frac{\sqrt[]{5} + 1}{4}] \\ = \frac{\sqrt[]{5} - 1 - \sqrt[]{5} - 1}{4} = \frac{- 1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

If $t a n A = \frac{1}{2}$ and $t a n B = \frac{1}{3}$ , then $t a n (2 A + B)$ is equal to:

(a) 1

(b) 2

(c) 3

(d) 4

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n A = \frac{1}{2} a n d t a n B = \frac{1}{3} \\ t a n 2 A = \frac{2 t a n A}{1 - t a n^{2} A} = \frac{2 \times \frac{1}{2}}{1 - {(\frac{1}{2})}^{2}} \\ = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \\ S o, t a n 2 A = \frac{4}{3} a n d t a n B = \frac{1}{3} \\ t a n (2 A + B) = \frac{t a n 2 A + t a n B}{1 - t a n 2 A . t a n B} = \frac{\frac{4}{3} + \frac{1}{3}}{1 - \frac{4}{3} \times \frac{1}{3}} = \frac{\frac{5}{3}}{\frac{9 - 4}{9}} = \frac{5}{3} \times \frac{9}{5} = 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The value of $s i n \frac{π}{10} s i n \frac{13 π}{10}$ is:

(a) $\frac{1}{2}$

(b) $- \frac{1}{2}$

(c) $- \frac{1}{4}$

(d) 1

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ s i n \frac{π}{10} . s i n \frac{13 π}{10} = s i n \frac{π}{10} . s i n (π + \frac{3 π}{10}) \\ = s i n \frac{π}{10} . (- s i n \frac{3 π}{10}) = - sin 1 8^{0} . sin 5 4^{0} \\ = - sin 1 8^{0} . sin (9 0^{0} - 3 6^{0}) = - sin 1 8^{0} . c o s 3 6^{0} \\ = - (\frac{\sqrt[]{5} - 1}{4}) (\frac{\sqrt[]{5} + 1}{4}) [? sin 1 8^{0} = \frac{\sqrt[]{5} - 1}{4}, c o s 3 6^{0} = \frac{\sqrt[]{5} + 1}{4}] \\ = - (\frac{5 - 1}{16}) = - \frac{1}{4} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The value of $s i n 5 0^{\circ} - s i n 7 0^{\circ} + s i n 1 0^{\circ}$ is equal to:

(a) 1

(b) 0

(c) $- \frac{1}{4}$

(d) 1

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : sin 5 0^{0} - sin 7 0^{0} + sin 1 0^{0} \\ (sin 5 0^{0} - sin 7 0^{0}) + sin 1 0^{0} = 2 c o s \frac{5 0^{0} + 7 0^{0}}{2} . s i n \frac{5 0^{0} - 7 0^{0}}{2} + sin 1 0^{0} \\ = 2 c o s 6 0^{0} . (- sin 1 0^{0}) + sin 1 0^{0} \\ = - 2 \times \frac{1}{2} sin 1 0^{0} + sin 1 0^{0} \\ = - sin 1 0^{0} + sin 1 0^{0} = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If $s i n θ + c o s θ = 1$ , then the value of $s i n 2 θ$ is:

(a) 1

(b) $\frac{1}{2}$

(c) 0

(d) -1

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n θ + c o s θ = 1 \\ \Rightarrow {(s i n θ + c o s θ)}^{2} = {(1)}^{2} \\ \Rightarrow s i n^{2} θ + c o s^{2} θ + 2 s i n θ c o s θ = 1 \\ \Rightarrow 1 + s i n 2 θ = 1 \\ \Rightarrow s i n 2 θ = 1 - 1 = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

If $α + β = \frac{π}{4}$ , then the value of $(1 + t a n α) (1 + t a n β)$ is:

(a) 1

(b) 2

(c) -2

(d) Not defined

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : α + β = \frac{π}{4} \\ W e k n o w t h a t \\ t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α . t a n β} \\ \Rightarrow t a n α + t a n β = 1 - t a n α . t a n β \\ \Rightarrow t a n α + t a n β + t a n α . t a n β = 1 \\ \Rightarrow 1 + t a n α + t a n β + t a n α . t a n β = 1 + 1 \\ \Rightarrow 1 (1 + t a n α) + t a n β (1 + t a n α) = 2 \\ \Rightarrow (1 + t a n α) (1 + t a n β) = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If $s i n θ = - \frac{2}{5}$ and $θ$ lies in the third quadrant, then the value of $c o s θ$ is:

(a) $- \frac{1}{5}$

(b) $- \frac{\sqrt[]{21}}{5}$

(c) $\frac{1}{5}$

(d) $\frac{\sqrt[]{21}}{5}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n θ = - \frac{4}{5}, θ l i e s i n t h i r d q u a d r a n t \\ c o s θ = \sqrt[]{1 - s i n^{2} θ} = \sqrt[]{1 - {(- \frac{4}{5})}^{2}} = \sqrt[]{1 - \frac{16}{25}} = \sqrt[]{\frac{9}{25}} = \frac{+ 3}{- 5} \\ ∴ c o s θ = - \frac{3}{5}, θ l i e s i n t h i r d q u a d r a n t \\ c o s θ = 2 c o s^{2} \frac{θ}{2} - 1 [? π < θ < \frac{3 π}{2}, ∴ \frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4}] \\ \Rightarrow \frac{- 3}{5} = 2 c o s^{2} \frac{θ}{2} - 1 \\ \Rightarrow 2 c o s^{2} \frac{θ}{2} = 1 - \frac{3}{5} = \frac{2}{5} \Rightarrow c o s^{2} \frac{θ}{2} = \frac{2}{5 \times 2} = \frac{1}{5} \\ \Rightarrow c o s \frac{θ}{2} = \pm \frac{1}{\sqrt[]{5}} \Rightarrow c o s \frac{θ}{2} = - \frac{1}{\sqrt[]{5}} [? \frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Number of solutions of the equation $t a n x + s e c x = 2 c o s x$ lying in the interval $[0, 2 π]$ is:

(a) 0

(b) 1

(c) 2

(d) 3

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n x + s e c x = 2 c o s x \\ \Rightarrow \frac{s i n x}{c o s x} + \frac{1}{c o s x} = 2 c o s x \\ \Rightarrow 1 + s i n x = 2 c o s^{2} x \Rightarrow 2 c o s^{2} x - s i n x - 1 = 0 \\ \Rightarrow 2 (1 - si n^{2} x) - s i n x - 1 = 0 \Rightarrow 2 - 2 si n^{2} x - s i n x - 1 = 0 \\ \Rightarrow - 2 si n^{2} x - s i n x + 1 = 0 \Rightarrow 2 si n^{2} x + s i n x - 1 = 0 \\ Since t h e e q u a t i o n i s q u a d r a t i c e q u a t i o n i n s i n x . S o i t w i l l h a v e 2 s o l u t i o n s . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The value of $s i n \frac{2 π}{18} + s i n \frac{5 π}{18} + s i n \frac{7 π}{18} + s i n \frac{4 π}{9}$ is given by:

(a) 1

(b) -1

(c) $c o s \frac{3 π}{9} + s i n \frac{π}{18}$

(d) $c o s \frac{2 π}{9} + s i n \frac{π}{18}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s s i n \frac{π}{18} + s i n \frac{π}{9} + s i n \frac{2 π}{9} + s i n \frac{5 π}{18} \\ = (s i n \frac{5 π}{18} + s i n \frac{π}{18}) + (s i n \frac{2 π}{9} + s i n \frac{π}{9}) \\ = 2 s i n (\frac{\frac{5 π}{18} + \frac{π}{18}}{2}) . c o s (\frac{\frac{5 π}{18} - \frac{π}{18}}{2}) + 2 s i n (\frac{\frac{2 π}{9} + \frac{π}{9}}{2}) . c o s (\frac{\frac{2 π}{9} - \frac{π}{9}}{2}) \\ = 2 s i n \frac{π}{6} . c o s \frac{π}{9} + 2 s i n \frac{π}{6} . c o s \frac{π}{18} \\ = 2 \times \frac{1}{2} c o s \frac{π}{9} + 2 \times \frac{1}{2} c o s \frac{π}{18} = c o s \frac{π}{9} + c o s \frac{π}{18} \\ = s i n (\frac{π}{2} - \frac{π}{9}) + s i n (\frac{π}{2} - \frac{π}{18}) = s i n \frac{7 π}{18} + s i n \frac{8 π}{18} \\ = s i n \frac{7 π}{18} + s i n \frac{4 π}{9} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $A$ lies in the second quadrant and $3 t a n A + 4 = 0$ , then the value of $2 c o t A - 5 c o s A + s i n A$ is equal to:

(a) $- \frac{53}{10}$

(b) $- \frac{23}{10}$

(c) $- \frac{37}{10}$

(d) None

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : 3 t a n A + 4 = 0, A l i e s i n second q u a d r a n t \\ ∴ t a n A = \frac{- 4}{3} \\ c o s A = \frac{- 3}{5} [A l i e s i n second q u a d r a n t] \\ a n d s i n A = \frac{4}{5} a n d c o t A = \frac{- 3}{4} \\ ∴ 2 c o t A - 5 c o s A + s i n A = 2 (\frac{- 3}{4}) - 5 (\frac{- 3}{5}) + \frac{4}{5} \\ = \frac{- 3}{2} + 3 + \frac{4}{5} = \frac{- 15 + 30 + 8}{10} = \frac{23}{10} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The value of $c o s^{2} 4 8^{?} - s i n^{2} 1 2^{?}$ is:

(a) $\frac{\sqrt[]{5} + 1}{4}$

(b) $- \frac{\sqrt[]{5} + 1}{4}$

(c) $\frac{1}{2}$

(d) $- \frac{1}{2}$

Hint : Use of $c o s^{2} A - s i n^{2} B = c o s (A + B) C o s (A - B)$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s c o s^{2} 4 8^{0} - s i n^{2} 1 2^{0} \\ c o s^{2} 4 8^{0} - s i n^{2} 1 2^{0} = c o s (4 8^{0} + 1 2^{0}) . c o s (4 8^{0} - 1 2^{0}) [? c o s^{2} A - s i n^{2} B = c o s (A + B) . c o s (A - B)] \\ = c o s 6 0^{0} . c o s 3 6^{0} = \frac{1}{2} \times \frac{\sqrt[]{5} + 1}{4} = \frac{\sqrt[]{5} + 1}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $t a n α = \frac{1}{7}$ , $t a n β = \frac{1}{3}$ , then $c o s 2 α$ is equal to:

(a) $s i n 2 β$

(b) $s i n 4 β$

(c) $s i n 3 β$

(d) $c o s 2 β$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n α = \frac{1}{7} a n d t a n β = \frac{1}{3} \\ c o s 2 α = \frac{1 - t a n^{2} α}{1 + t a n^{2} α} = \frac{1 - {(\frac{1}{7})}^{2}}{1 + {(\frac{1}{7})}^{2}} = \frac{1 - \frac{1}{49}}{1 + \frac{1}{49}} = \frac{48}{50} = \frac{24}{25} \\ t a n 2 β = \frac{2 t a n β}{1 - t a n^{2} β} = \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4} \\ ∴ t a n 2 β = \frac{3}{4} \\ s i n 4 β = \frac{2 t a n 2 β}{1 + t a n^{2} 2 β} = \frac{2 \times \frac{3}{4}}{1 + {(\frac{3}{4})}^{2}} = \frac{\frac{3}{2}}{1 + \frac{9}{16}} = \frac{3}{2} \times \frac{16}{25} = \frac{24}{25} \\ c o s 2 α = s i n 4 β = \frac{24}{25} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If $t a n θ = \frac{a}{b}$ , then $b c o s 2 θ + a s i n 2 θ$ is equal to:

(a) a

(b) b

(c) $\frac{b}{a}$

(d) None

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ = \frac{a}{b} \\ b c o s 2 θ + a s i n 2 θ = b [\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}] + a [\frac{2 t a n θ}{1 + t a n^{2} θ}] \\ = b [\frac{1 - \frac{a^{2}}{b^{2}}}{1 + \frac{a^{2}}{b^{2}}}] + a [\frac{\frac{2 a}{b}}{1 + \frac{a^{2}}{b^{2}}}] \\ = b [\frac{b^{2} - a^{2}}{b^{2} + a^{2}}] + [\frac{\frac{2 a^{2}}{b}}{\frac{b^{2} + a^{2}}{b^{2}}}] \\ = \frac{b^{3} - a^{2} b}{b^{2} + a^{2}} + \frac{2 a^{2} b}{b^{2} + a^{2}} = \frac{b^{3} - a^{2} b + 2 a^{2} b}{b^{2} + a^{2}} \\ = \frac{b^{3} + a^{2} b}{b^{2} + a^{2}} = \frac{b (b^{2} + a^{2})}{b^{2} + a^{2}} = b \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If $c o s θ = \frac{x}{x + 1}$ , then $θ$ is:

(a) An acute angle

(b) A right angle

(c) An obtuse angle

(d) Not possible

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s θ = x + \frac{1}{x} \Rightarrow c o s θ = \frac{x^{2} + 1}{x} \\ \Rightarrow x^{2} + 1 = x c o s θ \Rightarrow x^{2} - x c o s θ + 1 = 0 \\ F o r r e a l v a l u e o f x, b^{2} - 4 a c \geq 0 \\ \Rightarrow {(- c o s θ)}^{2} - 4 \times 1 \times 1 \geq 0 \\ \Rightarrow c o s^{2} θ - 4 \geq 0 \Rightarrow c o s^{2} θ \geq 4 \\ \Rightarrow c o s θ \geq \pm 2 [- 1 \leq c o s θ \leq 1] \\ S o, t h e v a l u e o f θ i s n o t p o s s i b l e . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Trigonometric Functions Fill in the blanks Type Questions

1. The value of $\frac{s i n 5 0^{\circ}}{s i n 13 0^{\circ}}$ is ___.

Sol:

$\begin{array}{l} \frac{s i n 5 0^{0}}{s i n 13 0^{0}} = \frac{s i n 5 0^{0}}{s i n (18 0^{0} - 5 0^{0})} = \frac{s i n 5 0^{0}}{s i n 5 0^{0}} = 1 \\ H e n c e, t h e v a l u e o f f i l l e r i s 1 . \end{array}$

2. If $k = s i n \frac{π}{18} s i n \frac{5 π}{18} s i n \frac{7 π}{18}$ , then the numerical value of $k$ is ___.

Sol:

$\begin{array}{l} G i v e n t h a t : k = s i n (\frac{π}{18}) s i n (\frac{5 π}{18}) s i n (\frac{7 π}{18}) \\ \Rightarrow k = s i n 1 0^{0} . s i n 5 0^{0} . s i n 7 0^{0} \\ \Rightarrow k = s i n 1 0^{0} . s i n (9 0^{0} - 4 0^{0}) . s i n (9 0^{0} - 2 0^{0}) \\ \Rightarrow k = s i n 1 0^{0} . c o s 4 0^{0} . c o s 2 0^{0} \\ \Rightarrow k = s i n 1 0^{0} . \frac{1}{2} [2 c o s 4 0^{0} . c o s 2 0^{0}] \\ \Rightarrow k = s i n 1 0^{0} . \frac{1}{2} [c o s (4 0^{0} + 2 0^{0}) + c o s (4 0^{0} - 2 0^{0})] \\ \Rightarrow k = \frac{1}{2} s i n 1 0^{0} [c o s 6 0^{0} + c o s 2 0^{0}] \\ \Rightarrow k = \frac{1}{2} s i n 1 0^{0} [\frac{1}{2} + c o s 2 0^{0}] \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{2} s i n 1 0^{0} c o s 2 0^{0} \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{4} (2 s i n 1 0^{0} c o s 2 0^{0}) \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{4} [s i n (1 0^{0} + 2 0^{0}) + s i n (1 0^{0} - 2 0^{0})] \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{4} [s i n 3 0^{0} + s i n (- 1 0^{0})] \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{4} s i n 3 0^{0} - \frac{1}{4} s i n 1 0^{0} \\ \Rightarrow k = \frac{1}{4} s i n 3 0^{0} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} . \\ H e n c e, t h e v a l u e o f f i l l e r i s \frac{1}{8} . \end{array}$

Commonly asked questions

The value of $\frac{s i n 5 0^{\circ}}{s i n 13 0^{\circ}}$ is ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

If $k = s i n \frac{π}{18} s i n \frac{5 π}{18} s i n \frac{7 π}{18}$ , then the numerical value of $k$ is ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

If $t a n A = \frac{1 - c o s B}{s i n B}$ , then $t a n 2 A$ is ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n A = \frac{1 - c o s B}{s i n B} \\ t a n 2 A = \frac{2 t a n A}{1 - t a n^{2} A} = \frac{2 (\frac{1 - c o s B}{s i n B})}{1 - {(\frac{1 - c o s B}{s i n B})}^{2}} \\ = \frac{2 (\frac{2 s i n^{2} B / 2}{2 s i n B / 2 c o s B / 2})}{1 - {(\frac{2 s i n^{2} B / 2}{2 s i n B / 2 c o s B / 2})}^{2}} [\begin{array}{l} ? 1 - c o s B = 2 s i n^{2} B / 2 \\ s i n B = 2 s i n B / 2 c o s B / 2 \end{array}] \\ = \frac{2 (\frac{s i n B / 2}{c o s B / 2})}{1 - {(\frac{s i n B / 2}{c o s B / 2})}^{2}} = \frac{2 t a n B / 2}{1 - t a n^{2} B / 2} = t a n B \\ S o, t a n 2 A = t a n B \\ H e n c e, t h e v a l u e o f f i l l e r i s t a n B . \end{array}$

If $s i n x + c o s x = a$ , then:

i. $s i n^{6} x + c o s^{6} x =$ ___
ii. $∣ s i n x - c o s x ∣ =$ ___

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n x + c o s x = a \\ {(s i n x + c o s x)}^{2} = a^{2} \\ \Rightarrow s i n^{2} x + c o s^{2} x + 2 s i n x c o s x = a^{2} \\ \Rightarrow 1 + 2 s i n x c o s x = a^{2} \\ \Rightarrow s i n x c o s x = \frac{a^{2} - 1}{2} \dots (i) \\ (i) s i n^{6} x + c o s^{6} x = {(s i n^{2} x)}^{3} + {(c o s^{2} x)}^{3} \\ = {(s i n^{2} x + c o s^{2} x)}^{3} - 3 s i n^{2} x c o s^{2} x (s i n^{2} x + c o s^{2} x) \\ = {(1)}^{3} - 3 {(\frac{a^{2} - 1}{2})}^{2} . 1 = 1 - \frac{3 {(a^{2} - 1)}^{2}}{4} = \frac{1}{4} [4 - 3 {(a^{2} - 1)}^{2}] \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{1}{4} [4 - 3 {(a^{2} - 1)}^{2}] . \\ (i i) {| s i n x - c o s x |}^{2} = s i n^{2} x + c o s^{2} x - 2 s i n x c o s x \\ = 1 - 2 (\frac{a^{2} - 1}{2}) = 1 - (a^{2} - 1) = 1 - a^{2} + 1 \\ = 2 - a^{2} \\ ∴ | s i n x - c o s x | = \sqrt[]{2 - a^{2}} [? | s i n x - c o s x | > 0] \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \sqrt[]{2 - a^{2}} . \end{array}$

In a triangle $△ A B C$ with $∠ C = 9 0^{\circ}$ , the equation whose roots are $t a n A$ and $t a n B$ is ___.

Hint :If $A + B = 9 0^{0} \Rightarrow t a n A t a n B = 1$ , then $t a n A + t a n B = \frac{2}{S i n A}$ ___.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n a Δ A B C w i t h ∠ C = 9 0^{0} \\ S o, t h e e q u a t i o n w h o s e r o o t s a r e t a n A a n d t a n B i s \\ x^{2} - (t a n A + t a n B) x + t a n A . t a n B = 0 \\ A + B = 9 0^{0} [? ∠ C = 9 0^{0}] \\ \Rightarrow t a n (A + B) = t a n 9 0^{0} \\ \Rightarrow \frac{t a n A + t a n B}{1 - t a n A . t a n B} = \frac{1}{0} \\ \Rightarrow 1 - t a n A . t a n B = 0 \\ \Rightarrow t a n A . t a n B = 1 \dots (i) \\ N o w t a n A + t a n B = \frac{s i n A}{c o s A} + \frac{s i n B}{c o s B} \\ = \frac{s i n A c o s B + c o s A s i n B}{c o s A c o s B} \\ = \frac{s i n (A + B)}{c o s A c o s B} = \frac{s i n 9 0^{0}}{c o s A . c o s (9 0^{0} - A)} \\ = \frac{1}{c o s A s i n A} \\ ∴ t a n A + t a n B = \frac{2}{2 c o s A s i n A} = \frac{2}{s i n 2 A} \dots (i i) \\ F r o m (i) a n d (i i) w e g e t \\ x^{2} - (\frac{2}{s i n 2 A}) x + 1 = 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s x^{2} - (\frac{2}{s i n 2 A}) x + 1 = 0 . \end{array}$

$3 {(s i n x - c o s x)}^{4} + 6 {(s i n x + c o s x)}^{2} + 4 (s i n^{6} x + c o s^{6} x) =$ ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s \\ 3 {(s i n x - c o s x)}^{4} + 6 {(s i n x + c o s x)}^{2} + 4 (s i n^{6} x + c o s^{6} x) \\ = 3 {[s i n^{2} x + c o s^{2} x - 2 s i n x c o s x]}^{2} + 6 [s i n^{2} x + c o s^{2} x + 2 s i n x c o s x] + 4 [{(s i n^{2} x)}^{3} + {(c o s^{2} x)}^{3}] \\ = 3 {[1 - 2 s i n x c o s x]}^{2} + 6 [1 + 2 s i n x c o s x] + 4 [{(s i n^{2} x + c o s^{2} x)}^{3} - 3 s i n^{2} x c o s^{2} x (s i n^{2} x + c o s^{2} x)] \\ = 3 [1 + 4 s i n^{2} x c o s^{2} x - 4 s i n x c o s x] + 6 (1 + 2 s i n x c o s x) + 4 [1 - 3 s i n^{2} x c o s^{2} x] \\ = 3 + 12 s i n^{2} x c o s^{2} x - 12 s i n x c o s x + 6 + 12 s i n x c o s x + 4 - 12 s i n^{2} x c o s^{2} x \\ = 3 + 6 + 4 = 13 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 13 . \end{array}$

Given $x > 0$ , the values of $f (x) = - 3 c o s 2 x + \frac{3}{x}$ lie in the interval ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : f (x) = - 3 c o s \sqrt[]{3 + x + x^{2}} \\ P u t \sqrt[]{3 + x + x^{2}} = y \\ ∴ f (x) = - 3 c o s y \\ ? - 1 \leq c o s y \leq 1 \\ 3 \geq - 3 c o s y \geq - 3 \\ \Rightarrow - 3 \leq - 3 c o s y \leq 3 \\ \Rightarrow - 3 \leq - 3 c o s \sqrt[]{3 + x + x^{2}} \leq 3, x > 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s [- 3, 3] . \end{array}$

The maximum distance of a point on the graph of the function $y = 3 s i n x + c o s x$ from the $x$ -axis is ___.

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : y = \sqrt[]{3} s i n x + c o s x \dots (i) \\ ∴ T h e maximum distance f r o m a point o n t h e g r a p h o f e q n . (i) f r o m x - a x i s \\ = \sqrt[]{{(\sqrt[]{3})}^{2} + {(1)}^{2}} = \sqrt[]{3 + 1} = 2 . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 2 . \end{array}$

Trigonometric Functions True or False Type Questions

1. If $t a n A = \frac{1 - c o s B}{s i n B}$ , then $t a n 2 A = t a n B$ .

Sol:

$\begin{array}{l} G i v e n t h a t : t a n A = \frac{1 - c o s B}{s i n B} = \frac{2 s i n^{2} B / 2}{2 s i n B / 2 c o s B / 2} = t a n \frac{B}{2} \\ t a n 2 A = \frac{2 t a n A}{1 - t a n^{2} A} = \frac{2 t a n B / 2}{1 - t a n^{2} B / 2} \\ ∴ t a n 2 A = t a n B \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

2. The equality $s i n A + s i n 2 A + s i n 3 A = 3$ holds for some real value of $A$ .

Sol:

$\begin{array}{l} G i v e n t h a t : s i n A + s i n 2 A + s i n 3 A = 3 \\ Since t h e maximum v a l u e o f s i n A i s 1 b u t f o r s i n 2 A a n d s i n 3 A i t i s n o t e q u a l t o 1 . \\ S o i t i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

Commonly asked questions

If $t a n A = \frac{1 - c o s B}{s i n B}$ , then $t a n 2 A = t a n B$ .

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

The equality $s i n A + s i n 2 A + s i n 3 A = 3$ holds for some real value of $A$ .

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$s i n 1 0^{\circ}$ is greater than $c o s 1 0^{\circ}$ .

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I f s i n 1 0^{0} > c o s 1 0^{0} \\ \Rightarrow s i n 1 0^{0} > c o s (9 0^{0} - 8 0^{0}) \\ \Rightarrow s i n 1 0^{0} > s i n 8 0^{0} w h i c h i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

. $c o s \frac{π}{15} \cdot c o s \frac{4 π}{15} \cdot c o s \frac{8 π}{15} \cdot c o s \frac{16 π}{15} = \frac{1}{16}$

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L . H . S . c o s \frac{2 π}{15} . c o s \frac{4 π}{15} . c o s \frac{8 π}{15} . c o s \frac{16 π}{15} \\ = c o s 2 4^{0} . c o s 4 8^{0} . c o s 9 6^{0} . c o s 19 2^{0} \\ = \frac{1}{16 s i n 2 4^{0}} [(2 s i n 2 4^{0} c o s 2 4^{0}) (2 c o s 4 8^{0}) (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 4 8^{0} . 2 c o s 4 8^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 4 8^{0} c o s 4 8^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 9 6^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 9 6^{0} c o s 9 6^{0} (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 19 2^{0} (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 19 2^{0} c o s 19 2^{0}] \\ = \frac{1}{16 s i n 2 4^{0}} s i n 38 4^{0} = \frac{1}{16 s i n 2 4^{0}} s i n (36 0^{0} + 2 4^{0}) \\ = \frac{1}{16 s i n 2 4^{0}} \times s i n 2 4^{0} [? s i n (36 0^{0} + θ) = s i n θ] \\ = \frac{1}{16} R . H . S . \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

One value of $θ$ which satisfies the equation $s i n^{4} θ - 2 s i n^{2} θ - 1 = 0$ lies between $0$ and $2 π$ .

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s \\ s i n^{4} θ - 2 s i n^{2} θ - 1 = 0 \\ s i n^{2} θ = \frac{- (- 2) \pm \sqrt[]{{(- 2)}^{2} - 4 \times 1 \times - 1}}{2 \times 1} \\ = \frac{2 \pm \sqrt[]{4 + 4}}{2} = \frac{2 \pm \sqrt[]{8}}{2} = \frac{2 \pm 2 \sqrt[]{2}}{2} = 1 \pm \sqrt[]{2} \\ ∴ s i n^{2} θ = (1 + \sqrt[]{2}) o r (1 - \sqrt[]{2}) \Rightarrow - 1 \leq s i n θ \leq 1 \\ \Rightarrow s i n^{2} θ \leq 1 b u t s i n^{2} θ = (1 + \sqrt[]{2}) o r (1 - \sqrt[]{2}) \\ W h i c h i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

If $c o s e c x = 1 + c o t x$ , then $x = 2 n π, 2 n π + \frac{π}{2}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s e c x = 1 + c o t x \\ \Rightarrow \frac{1}{s i n x} = 1 + \frac{c o s x}{s i n x} \\ \Rightarrow \frac{1}{s i n x} = \frac{s i n x + c o s x}{s i n x} \\ \Rightarrow s i n x + c o s x = 1 \\ \Rightarrow \frac{1}{\sqrt[]{2}} s i n x + \frac{1}{\sqrt[]{2}} c o s x = \frac{1}{\sqrt[]{2}} \\ \Rightarrow s i n \frac{π}{4} s i n x + c o s \frac{π}{4} c o s x = \frac{1}{\sqrt[]{2}} \\ \Rightarrow c o s (x - \frac{π}{4}) = \frac{1}{\sqrt[]{2}} \\ \Rightarrow c o s (x - \frac{π}{4}) = c o s \frac{π}{4} \\ \Rightarrow x = 2 n π + \frac{π}{4} + \frac{π}{4} \Rightarrow x = 2 n π + \frac{π}{2} \\ o r x = 2 n π + \frac{π}{4} - \frac{π}{4} \Rightarrow x = 2 n π \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

If $t a n θ + t a n 2 θ + 3 t a n θ t a n 2 θ = 3$ , then $θ = \frac{n π}{3} + \frac{π}{9}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ + t a n 2 θ + \sqrt[]{3} t a n θ t a n 2 θ = \sqrt[]{3} \\ \Rightarrow t a n θ + t a n 2 θ = \sqrt[]{3} - \sqrt[]{3} t a n θ t a n 2 θ \\ \Rightarrow t a n θ + t a n 2 θ = \sqrt[]{3} (1 - t a n θ t a n 2 θ) \\ \Rightarrow \frac{t a n θ + t a n 2 θ}{1 - t a n θ t a n 2 θ} = \sqrt[]{3} \\ \Rightarrow t a n (θ + 2 θ) = \sqrt[]{3} \\ \Rightarrow t a n 3 θ = t a n \frac{π}{3} ∴ 3 θ = n π + \frac{π}{3} \\ S o, θ = \frac{n π}{3} + \frac{π}{9} \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

If $t a n (π c o s θ) = c o t (π s i n θ)$ , then $c o s (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt[]{2}}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n (π c o s θ) = c o t (π s i n θ) \\ \Rightarrow t a n (π c o s θ) = t a n (\frac{π}{2} - π s i n θ) \\ \Rightarrow π c o s θ = \frac{π}{2} - π s i n θ \\ \Rightarrow π c o s θ + π s i n θ = \frac{π}{2} \\ \Rightarrow c o s θ + s i n θ = \frac{1}{2} \\ \Rightarrow \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ = \frac{1}{2 \sqrt[]{2}} \\ \Rightarrow c o s \frac{π}{4} c o s θ + s i n \frac{π}{4} s i n θ = \frac{1}{2 \sqrt[]{2}} \\ \Rightarrow c o s (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt[]{2}} [? c o s (θ - \frac{π}{4}) o r c o s (\frac{π}{4} - θ)] \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

Match the following items under column to their correct answers under column :

i. $s i n (x + y) s i n (x - y)$ (i) $c o s^{2} x - s i n^{2} y$

ii. $c o s (x + y) c o s (x - y)$ (ii) $\frac{1 - t a n θ}{1 + t a n θ}$

iii. $c o t (\frac{π}{4} + θ)$ (iii) $\frac{1 + t a n θ}{1 - t a n θ}$

iv. $t a n (\frac{π}{4} + θ)$ (iv) $s i n^{2} x - s i n^{2} y$

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (a) s i n (x + y) s i n (x - y) = s i n^{2} x - s i n^{2} y \\ ∴ (a) \leftrightarrow (i v) \\ (b) c o s (x + y) c o s (x - y) = c o s^{2} x - c o s^{2} y \\ ∴ (b) \leftrightarrow (i) \\ (c) c o t (\frac{π}{4} + θ) = \frac{c o t \frac{π}{4} c o t θ - 1}{c o t θ + c o t \frac{π}{4}} = \frac{c o t θ - 1}{c o t θ + 1} = \frac{1 - t a n θ}{1 + t a n θ} \\ ∴ (c) \leftrightarrow (i i) \\ (d) t a n (\frac{π}{4} + θ) = \frac{t a n \frac{π}{4} + t a n θ}{1 - t a n \frac{π}{4} t a n θ} = \frac{1 + t a n θ}{1 - t a n θ} \\ ∴ (d) \leftrightarrow (i i i) \\ H e n c e, (a) \leftrightarrow (i v), (b) \leftrightarrow (i), (c) \leftrightarrow (i i) a n d (d) \leftrightarrow (i i i) . \end{array}$

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