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 .

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 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 ) .

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 ' .

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

Q: 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: 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 ⇒ s i n 2 θ + c o s e c 2 θ + 2 s i n θ c o s e c θ = 4 ⇒ s i n 2 θ + c o s e c 2 θ + 2 s i n θ × 1 s i n θ = 4 ⇒ s i n 2 θ + c o s e c 2 θ + 2 = 4 ⇒ 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 ) .

Updated on Aug 29, 2025 13:59 IST

By alok kumar singh, Executive Content Operations

Table of content

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

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 $\frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 + c o s α + s i n α}{1 + c o s α + s i n α}$

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.]

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 (( $θ +$ $α$ ) $-$ $(θ + β)$ )]

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.]

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

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

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.]

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) $\frac{π}{4}$

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 ___.

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 ___.

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

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 $t a n 2 A = t a n B$ .

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

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

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