# NCERT SOLUTIONS FOR CLASS 11 MATHS CHAPTER 3 / TRIGONOMETRIC FUNCTIONS EX 3.3 /

Question 1. Prove that: ${ sin }^{ 2 }\frac { \pi }{ 6 } +{ cos }^{ 2 }\frac { \pi }{ 3 } -{ tan }^{ 2 }\frac { \pi }{ 4 } =-\frac { 1 }{ 2 }$

Solution.

L.H.S. = ${ sin }^{ 2 }\frac { \pi }{ 6 } +{ cos }^{ 2 }\frac { \pi }{ 3 } -{ tan }^{ 2 }\frac { \pi }{ 4 } =-\frac { 1 }{ 2 }$

$=\left[ { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( 1 \right) }^{ 2 } \right] =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } -1=\frac { -1 }{ 2 } =\quad R.H.S.$

Question 2. $2{ sin }^{ 2 }\frac { \pi }{ 6 } +{ cosec }^{ 2 }\frac { 7\pi }{ 6 } { cos }^{ 2 }\frac { \pi }{ 3 } =\frac { 3 }{ 2 }$

Solution.

L.H.S. = $2{ sin }^{ 2 }\frac { \pi }{ 6 } +{ cosec }^{ 2 }\frac { 7\pi }{ 6 } { cos }^{ 2 }\frac { \pi }{ 3 }$

Question 3. ${ cot }^{ 2 }\frac { \pi }{ 6 } +cosec\frac { 5\pi }{ 6 } +3{ tan }^{ 2 }\frac { \pi }{ 6 } =6$

Solution.

L.H.S. = ${ cot }^{ 2 }\frac { \pi }{ 6 } +cosec\frac { 5\pi }{ 6 } +3{ tan }^{ 2 }\frac { \pi }{ 6 }$

Question 4. $2{ sin }^{ 2 }\frac { 3\pi }{ 4 } +2{ cos }^{ 2 }\frac { \pi }{ 4 } +2{ sec }^{ 2 }\frac { \pi }{ 3 } =10$

Solution.

L.H.S. = $2{ sin }^{ 2 }\frac { 3\pi }{ 4 } +2{ cos }^{ 2 }\frac { \pi }{ 4 } +2{ sec }^{ 2 }\frac { \pi }{ 3 }$

Question 5. Find the value of:

• (i) sin 75°
• (ii) tan 15°

Solution.

(i) sin (75°) = sin (30° + 45°)

(ii) tan 15° = tan (45° – 30°)

Prove the following:

Question 6. $cos\left( \frac { \pi }{ 4 } -x \right) cos\left( \frac { \pi }{ 4 } -y \right) -sin\left( \frac { \pi }{ 4 } -x \right) sin\left( \frac { \pi }{ 4 } -y \right)$

Solution.

We have,

Question 7.

$\frac { tan\left( \frac { \pi }{ 4 } +x \right) }{ tan\left( \frac { \pi }{ 4 } -x \right) } ={ \left( \frac { 1+tan\quad x }{ 1-tan\quad x } \right) }^{ 2 }$

Solution.

We have,

Question 8.

$\frac { cos\left( \pi +x \right) cos\left( -x \right) }{ sin\left( \pi -x \right) cos\left( \frac { \pi }{ 2 } +x \right) } ={ cot }^{ 2 }x$

Solution.

We have,

Question 9.

$cos\left( \frac { 3\pi }{ 2 } +x \right) cos\left( 2\pi +x \right) \left[ cot\left( \frac { 3\pi }{ 2 } -x \right) +cot\left( 2\pi +x \right) \right] =1$

Solution.

We have,

Question 10. sin(n +1 )x sin(n + 2)x + cos(n +1 )x cos(n + 2)x = cosx

Solution.

We have,

Question 11.

$cos\left( \frac { 3\pi }{ 4 } +x \right) -cos\left( \frac { 3\pi }{ 4 } -x \right) =-\sqrt { 2 } sinx$

Solution.

We have,

Question 12. sin26x – sin24x= sin2x sin10x

Solution.

Question 13. cos22x – cos26x = sin 4x sin 8x

Solution.

Question 14. sin2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x

Solution.

We have,

Question 15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution.

Question 16.

$\frac { cos9x-cos5x }{ sin17x-sin3x } =-\frac { sin2x }{ cos10x }$

Solution.

We have,

Question 17.

$\frac { sin5x+sin3x }{ cos5x+cos3x } =tan4x$

Solution.

We have,

Question 18.

$\frac { sinx-siny }{ cosx+cosy } =tan\left( \frac { x-y }{ 2 } \right)$

Solution.

Question 19.

$\frac { sinx+sin3x }{ cosx+cos3x } =tan2x$

Solution.

Question 20.

$\frac { sinx-sin3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } =2sinx$

Solution.

Question 21.

$\frac { cos4x+cos3x+cos2x }{ sin4x+sin3x+sin2x } =cot3x$

Solution.

Question 22. cot x cot 2x – cot 2x cot 3x – cot3x cotx = 1

Solution.

We know that 3x = 2x + x.

Therefore,

Question 23.

$tan4x=\frac { 4tanx\left( 1-{ tan }^{ 2 }x \right) }{ 1-6{ tan }^{ 2 }x+{ tan }^{ 4 }x }$

Solution.

Question 24. cos 4x = 1 – 8 sin2x cos2x

Solution.

Question 25. cos 6x = 32 cos6 x – 48 cos4x + 18 cos2 x -1

Solution.