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Ncert Solutions Maths class 11th

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Ncert Solutions Maths class 11th Trigonometric Functions

22. Find the value of:

(i) sin 75° (ii) tan 15°

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Payal Gupta

Contributor-Level 10

22. (i) sin 75°= sin (45°+30°)

Using sin (x + y)= sin x cos y + cos x sin y we can write

sin 75°

= sin 45°cos 30°+ 45° sin 30°

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Ncert Solutions Maths class 11th Trigonometric Functions

21. $2 s i n^{2} \frac{3 π}{4} + 2 c o s^{2} \frac{π}{4} + 2 s e c^{2} \frac{π}{3} = 10$

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Payal Gupta

Contributor-Level 10

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Ncert Solutions Maths class 11th Trigonometric Functions

20. $c o t^{2} \frac{π}{6} + c o s e c \frac{5 π}{6} + 3 t a n^{2} \frac{π}{6} = 6$

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Payal Gupta

Contributor-Level 10

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Ncert Solutions Maths class 11th Trigonometric Functions

19. $2 S i n^{2} \frac{π}{6} + c o s e c^{2} \frac{7 π}{6} c o s^{2} \frac{π}{3} = \frac{3}{2}$

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Payal Gupta

Contributor-Level 10

19.

$= \frac{1}{2} + 4 . \frac{1}{4}$

$= \frac{1}{2} + 1$ $= \frac{3}{2}$

= R.H.S.

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Ncert Solutions Maths class 11th Trigonometric Functions

Class 11 Math Trigonometric Functions Exercise 3.3 Solution

18. $s i n^{2} \frac{π}{6} + c o s^{2} \frac{π}{3} - t a n^{2} \frac{π}{4} = - \frac{1}{2}$

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Payal Gupta

Contributor-Level 10

18.

$\begin{array}{l} {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} - 1^{2} \\ = \frac{1}{4} + \frac{1}{4} - 1 \\ = \frac{1 + 1 + 4}{4} \\ = \frac{- 2}{4} \\ = \frac{- 1}{2} \end{array}$