Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.3 | Set 1

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Prove that:

Question 1: sin² $\frac{\pi}{6}$ +cos² $\frac{\pi}{3}$ – tan² $\frac{\pi}{4}$ = $-\frac{1}{2}$

Solution:

Taking LHS in consideration, we get

= sin² $\frac{\pi}{6}$ +cos² $\frac{\pi}{3}$ – tan² $\frac{\pi}{4}$

Substituting the values,

= $(\frac{1}{2})^2 + (\frac{1}{2})^2 - (1)^2$

= $\frac{1}{2}$ – 1

= $-\frac{1}{2}$

Hence, LHS = RHS

Question 2: 2sin² $\frac{\pi}{6}$ +cosec² $\frac{7\pi}{6}$ cos² $\frac{\pi}{3}$ = $\frac{3}{2}$

Solution:

Taking LHS in consideration, we get

= 2sin² $\frac{\pi}{6}$ +cosec² $(\pi + \frac{\pi}{6})$ cos² $\frac{\pi}{3}$

= 2sin² $\frac{\pi}{6}$ + (- cosec $\frac{\pi}{6})^2$ cos² $\frac{\pi}{3}$

Substituting the values,

= $2(\frac{1}{2})^2 + ((-2)^2) (\frac{1}{2})^2$

= 2 $(\frac{1}{4})$ + 4 $(\frac{1}{4})$

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

= $\frac{3}{2}$

Hence, LHS = RHS

Question 3: cot² $\frac{\pi}{6}$ + cosec $\frac{5\pi}{6}$ + 3tan² $\frac{\pi}{6}$ = 6

Solution:

Taking LHS in consideration, we get

= cot² $\frac{\pi}{6}$ + cosec $(\pi - \frac{\pi}{6})$ + 3tan² $\frac{\pi}{6}$

= cot² $\frac{\pi}{6}$ + cosec $\frac{\pi}{6}$ + 3tan² $\frac{\pi}{6}$

Substituting the values,

= (√3)² + 2 + 3 $(\frac{1}{\sqrt{3}})^2$

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

= 6

Hence, LHS = RHS

Question 4: 2sin² $\frac{3\pi}{4}$ + 2cos² $\frac{\pi}{4}$ + 2sec² $\frac{\pi}{3}$ = 10

Solution:

Taking LHS in consideration, we get

= 2sin² $(\pi - \frac{\pi}{4})$ + 2cos² $\frac{\pi}{4}$ + 2sec² $\frac{\pi}{3}$

= 2sin² $\frac{\pi}{4}$ + 2cos² $\frac{\pi}{4}$ + 2sec² $\frac{\pi}{3}$

Substituting the values,

= $2(\frac{1}{\sqrt{2}})^2 + 2(\frac{1}{\sqrt{2}})^2 + 2(2)^2$

= 2 $(\frac{1}{2}) + 2(\frac{1}{2})$ + 2(4)

= 1 + 1 + 8

= 10

Hence, LHS = RHS

Question 5: Find the value of:

(i) sin 75°

Solution:

As, we don’t know the angle value for 75°, so we will break into the angles which we know.

75° = 30° + 45°, so lets use this and solve for sin(30° + 45°)

Using the trigonometric formula,

sin (A+B) = sin A cos B + cos A sin B

sin(30° + 45°) = sin 30° cos 45° + cos 30° sin 45°

Substituting values, we get

sin(75°) = $(\frac{1}{2}) (\frac{1}{\sqrt{2}}) + (\frac{\sqrt{3}}{2}) (\frac{1}{\sqrt{2}})$

sin(75°) = $(\frac{1}{2\sqrt{2}}) + (\frac{\sqrt{3}}{2\sqrt{2}})$

sin(75°) = $(\frac{1+\sqrt{3}}{2\sqrt{2}})$

(ii) tan 15°

Solution:

As, we don’t know the angle value for 15°, so we will break into the angles which we know.

15° = 60° – 45° or 45° – 30° so lets use this and solve for tan(45° – 30°)

Using the trigonometric formula,

tan (A-B) = $\mathbf{\frac{tan A-tan B}{1+tan A tan B}}$

tan(45° – 30°) = $\frac{tan 45\degree-tan 30\degree}{1+tan 45\degree tan 30\degree}$

Substituting values, we get

tan(15°) = $\frac{1-\frac{1}{\sqrt{3}}}{1+(1)(\frac{1}{\sqrt{3}})}$

tan(15°) = $\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$

tan(15°) = $\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Now rationalizing the denominator, multiply and divide by $(\sqrt{3}-1)$

tan(15°) = $\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$

tan(15°) = $\frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-1^2}$

tan(15°) = $\frac{(\sqrt{3})^2+1^2 - 2(\sqrt{3})(1)}{3-1}$

tan(15°) = $\frac{3+1 - 2\sqrt{3}}{2}$

tan(15°) = $\frac{4 - 2\sqrt{3}}{2}$

tan(15°) = 2 – $\sqrt{3}$

Prove the following:

Question 6: $cos (\frac{\pi}{4}-x)cos (\frac{\pi}{4}-y)- sin (\frac{\pi}{4}-x)sin (\frac{\pi}{4}-y)$ = sin (x+y)

Solution:

Taking LHS in consideration, we get

= $cos (\frac{\pi}{4}-x)cos (\frac{\pi}{4}-y)- sin (\frac{\pi}{4}-x)sin (\frac{\pi}{4}-y)$

As, here there is multiplication of cos cos and sin sin, we will use Defactorisation Formulae,

2 cos A cos B = cos (A+B) + cos (A-B) and, ……………….(1)

2 sin A sin B = cos (A-B) – cos (A+B) ……………….(2)

Multiply and divide LHS by 2, we get

= $\frac{2}{2}(cos (\frac{\pi}{4}-x)cos (\frac{\pi}{4}-y)- sin (\frac{\pi}{4}-x)sin (\frac{\pi}{4}-y))$

= $\frac{1}{2}(2 cos (\frac{\pi}{4}-x)cos (\frac{\pi}{4}-y)- 2 sin (\frac{\pi}{4}-x)sin (\frac{\pi}{4}-y))$

Subtracting (2) from (1) and using identity formulae, we get

2 cos A cos B – 2 sin A sin B = cos (A+B) + cos (A-B) – (cos (A-B) – cos (A+B))

2 cos A cos B – 2 sin A sin B = 2 cos (A+B)

Hence, using this

= $\frac{1}{2}[2 cos ((\frac{\pi}{4}-x)+ (\frac{\pi}{4}-y))]$

= cos $(\frac{2\pi}{4}-x-y)$

= cos $(\frac{\pi}{2}-(x+y))$

= sin (x+y) (As cos $(\frac{\pi}{2}-\theta)$ = sin θ)

Hence, LHS = RHS

Question 7: $\frac{tan(\frac{\pi}{4}+x)}{tan(\frac{\pi}{4}-x)} = (\frac{1+tan x}{1-tan x})^2$

Solution:

Taking LHS in consideration, we get

$\frac{tan(\frac{\pi}{4}+x)}{tan(\frac{\pi}{4}-x)}$

As, using Factorisation Formulae of tan, we have

tan (A+B) = $\mathbf{\frac{tan A + tan B}{1 - tan A tan B}}$ and,

tan (A-B) = $\mathbf{\frac{tan A - tan B}{1 + tan A tan B}}$

Now, substituting the values

= $\frac{\frac{tan (\frac{\pi}{4}) + tan x}{1 - tan (\frac{\pi}{4}) tan x}}{\frac{tan (\frac{\pi}{4}) - tan x}{1 + tan (\frac{\pi}{4}) tan x}}$

= $\frac{\frac{1 + tan x}{1 - (1) tan x}}{\frac{1 - tan x}{1 + (1) tan x}}$

= $\frac{1 + tan x}{1 - tan x} \times \frac{1 + tan x}{1 - tan x}$

= $(\frac{1 + tan x}{1 - tan x})^2$

Hence, LHS = RHS

Question 8: $\frac{cos(\pi + x)\hspace{0.1cm}cos(-x)}{sin (\pi-x) \hspace{0.1cm}cos(\frac{\pi}{2}+x)}$ = cot²x

Solution:

Taking LHS in consideration, we get

= $\frac{cos(\pi + x)\hspace{0.1cm}cos(-x)}{sin (\pi-x) \hspace{0.1cm}cos(\frac{\pi}{2}+x)}$

As, we know these standard values

cos(-x) = cos x

cos $(\pi + x)$ = – cos x

sin $(\pi-x)$ = sin x

cos $(\frac{\pi}{2} + x)$ = – sin x

Substituting these values, we have

= $\frac{(-cos x)\hspace{0.1cm}cos x}{sin x \hspace{0.1cm}(- sin x)}$

= $\frac{cos^2 x}{sin^2 x}$

= cot² x

Hence, LHS = RHS

Question 9: $cos (\frac{3\pi}{2}+x)\hspace{0.1cm}cos (2\pi+x)[\hspace{0.1cm}cot (\frac{3\pi}{2}-x)+cot(2\pi+x)]$ = 1

Solution:

Taking LHS in consideration, we get

= $cos (\frac{3\pi}{2}+x)\hspace{0.1cm}cos (2\pi+x)[\hspace{0.1cm}cot (\frac{3\pi}{2}-x)+cot(2\pi+x)]$

As, we know these standard values

cos $(\frac{3\pi}{2}+x)$ = sin x

cos $(2\pi + x)$ = cos x

cot $(2\pi + x)$ = cot x

cot $(\frac{3\pi}{2}-x)$ = tan x

Substituting the values, we have

= $sin x\hspace{0.1cm}cos x[\hspace{0.1cm}tan x + cot x]$

= $sin x\hspace{0.1cm}cos x[\frac{sin x}{cos x} + \frac{cos x}{sin x}]$

= $sin x\hspace{0.1cm}cos x[\frac{sin^2 x + cos^2 x}{sin x \hspace{0.1cm}cos x}]$

As sin² x + cos² x = 1

= 1

Hence, LHS = RHS

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

Solution:

Taking LHS in consideration, we get

= sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x

As, here there is multiplication of cos cos and sin sin, we will use Defactorisation Formulae,

2 cos A cos B = cos (A+B) + cos (A-B) and, ……………….(1)

2 sin A sin B = cos (A-B) – cos (A+B) ……………….(2)

Multiply and divide LHS by 2, we get

= $\frac{2}{2}$ (sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x)

= $\frac{1}{2}$ (2 sin(n + 1)x sin(n + 2)x + 2 cos(n + 1)x cos(n + 2)x)

Adding (1) and (2) and using identity formulae, we get

2 cos A cos B + 2 sin A sin B = cos (A+B) + cos (A-B) + cos (A-B) – cos (A+B)

2 cos A cos B + 2 sin A sin B = 2 cos (A-B)

Hence, using this

= $\frac{1}{2}$ (2 cos((n + 1)x – (n + 2)x))

= cos((n + 1)x – (n + 2)x)

= cos (x-2x)

= cos (- x)

= cos x (As, cos(-x) = cos x)

Hence, LHS = RHS

Question 11: $cos (\frac{3\pi}{4}+x)-cos (\frac{3\pi}{4}-x)$ = – $\sqrt{2}$ sin x

Solution:

Taking LHS in consideration, we get

= $cos (\frac{3\pi}{4}+x)-cos (\frac{3\pi}{4}-x)$

Using the identity,

cos A – cos B = 2 sin $\mathbf{(\frac{A+B}{2})}$ sin $\mathbf{(\frac{B-A}{2})}$

Substituting the values,

= $2 sin (\frac{(\frac{3\pi}{4}-x)+(\frac{3\pi}{4}+x)}{2}) sin (\frac{(\frac{3\pi}{4}-x)-(\frac{3\pi}{4}+x)}{2})$

= 2 sin $(\frac{6\pi}{4\times 2})$ sin $(\frac{(-x-x)}{2})$

= 2 sin $(\frac{3\pi}{4})$ sin (-x)

= 2 sin $(\pi - \frac{\pi}{4})$ sin (-x)

= 2 ( sin $(\frac{\pi}{4})$ ) sin (-x)

= 2 $(\frac{1}{\sqrt{2}})$ (- sin x)

= $\frac{-2 sin x}{\sqrt{2}}$

Rationalizing the denominator, by multiplying and dividing by $\sqrt{2}$

= $\frac{-2 sin x}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

= $\frac{-2 \sqrt{2} \hspace{0.1cm}sin x}{(\sqrt{2})^2}$

= $\frac{-2 \sqrt{2}\hspace{0.1cm} sin x}{2}$

= $-\sqrt{2}$ sin x

Hence, LHS = RHS

Question 12: sin² 6x – sin² 4x = sin 2x sin 10x

Solution:

Taking LHS in consideration, we get

= sin² 6x – sin² 4x

= sin 6x sin 6x – sin 4x sin 4x

As, here there is multiplication of sin sin, we will use Defactorisation Formulae,

2 sin A sin B = cos (A-B) – cos (A+B)

Multiply and divide LHS by 2, we get

= $\frac{2}{2}$ (sin 6x sin 6x – sin 4x sin 4x)

= $\frac{1}{2}$ (2 sin 6x sin 6x – 2 sin 4x sin 4x)

Using the identity, we can simplify

= $\frac{1}{2}$ [(cos(6x-6x) – cos(6x+6x)) – (cos(4x-4x) – cos(4x+4x))]

= $\frac{1}{2}$ [(cos(0) – cos(12x)) – (cos(0) – cos(8x))]

= $\frac{1}{2}$ [1 – cos(12x) – 1 + cos(8x)] (As, cos 0 = 1)

= $\frac{1}{2}$ [cos(8x) – cos (12x)]

Now, using the identity

cos A – cos B = 2 sin $\mathbf{(\frac{A+B}{2})}$ sin $\mathbf{(\frac{B-A}{2})}$

Substituting the values, we have

= $\frac{1}{2}[2 \hspace{0.1cm}sin (\frac{8x+12x}{2}) \hspace{0.1cm}sin (\frac{12x-8x}{2})]$

= sin $(\frac{20x}{2})$ sin $(\frac{4x}{2})$

= sin (10 x) sin (2x)

Hence, LHS = RHS

Question 13: cos² 2x – cos² 6x = sin 4x sin 8x

Solution:

Taking LHS in consideration, we get

= cos² 2x – cos² 6x

= cos 2x cos 2x – cos 6x cos 6x

As, here there is multiplication of cos cos, we will use Defactorisation Formulae,

2 cos A cos B = cos (A+B) + cos (A-B)

Multiply and divide LHS by 2, we get

= $\frac{2}{2}$ (cos 2x cos 2x – cos 6x cos 6x)

= $\frac{1}{2}$ (2 cos 2x cos 2x – 2 cos 6x cos 6x)

Using the identity, we can simplify

= $\frac{1}{2}$ [(cos(2x+2x) + cos(2x-2x)) – (cos(6x+6x) + cos(6x-6x))]

= $\frac{1}{2}$ [(cos(2x+2x) + cos(0)) – (cos(6x+6x) + cos(0))]

= $\frac{1}{2}$ [(cos(4x) + 1 – cos(12x) – 1)] (As, cos 0 = 1)

= $\frac{1}{2}$ [cos(4x) – cos(12x)]

Now, using the identity

cos A – cos B = 2 sin $\mathbf{(\frac{A+B}{2})}$ sin $\mathbf{(\frac{B-A}{2})}$

Substituting the values, we have

= $\frac{1}{2}[2 \hspace{0.1cm}sin (\frac{4x+12x}{2}) \hspace{0.1cm}sin (\frac{12x-4x}{2})]$

= sin $(\frac{16x}{2})$ sin $(\frac{8x}{2})$

= sin (8x) sin (4x)

Hence, LHS = RHS

Question 14: sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x

Solution:

Taking LHS in consideration, we get

sin 2x + 2 sin 4x + sin 6x

After rearranging, we have

= (sin 2x + sin 6x) + 2 sin 4x

Using the identity, we can simplify

sin A+ sin B = 2 sin $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

= 2 sin $(\frac{6x+2x}{2})$ cos $(\frac{6x-2x}{2})$ + 2 sin 4x

= 2 sin $(\frac{8x}{2})$ cos $(\frac{4x}{2})$ + 2 sin 4x

= 2 sin (4x) cos (2x) + 2 sin 4x

Taking (2 sin 4x), we have

= 2 sin (4x) (cos (2x) + 1)

= 2 sin (4x) (2 cos² x – 1 + 1) (As, cos 2θ = 2cos² θ – 1)

= 2 sin (4x) (2 cos²x)

= 4 sin (4x) cos² x

Hence, LHS = RHS

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Last Updated : 16 Apr, 2021

Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.3 | Set 1

Prove that:

Question 1: sin2+cos2– tan2=

Question 2: 2sin2+cosec2cos2=

Question 3: cot2+ cosec+ 3tan2= 6

Question 4: 2sin2+ 2cos2+ 2sec2= 10

Question 5: Find the value of:

(i) sin 75°

(ii) tan 15°

Prove the following:

Question 6:= sin (x+y)

Question 7:

Question 8:= cot2x

Question 9:= 1

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

Question 11:= –sin x

Question 12: sin2 6x – sin2 4x = sin 2x sin 10x

Question 13: cos2 2x – cos2 6x = sin 4x sin 8x

Question 14: sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x

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Question 1: sin² $\frac{\pi}{6}$ +cos² $\frac{\pi}{3}$ – tan² $\frac{\pi}{4}$ = $-\frac{1}{2}$

Question 2: 2sin² $\frac{\pi}{6}$ +cosec² $\frac{7\pi}{6}$ cos² $\frac{\pi}{3}$ = $\frac{3}{2}$

Question 3: cot² $\frac{\pi}{6}$ + cosec $\frac{5\pi}{6}$ + 3tan² $\frac{\pi}{6}$ = 6

Question 4: 2sin² $\frac{3\pi}{4}$ + 2cos² $\frac{\pi}{4}$ + 2sec² $\frac{\pi}{3}$ = 10

Question 6: $cos (\frac{\pi}{4}-x)cos (\frac{\pi}{4}-y)- sin (\frac{\pi}{4}-x)sin (\frac{\pi}{4}-y)$ = sin (x+y)

Question 7: $\frac{tan(\frac{\pi}{4}+x)}{tan(\frac{\pi}{4}-x)} = (\frac{1+tan x}{1-tan x})^2$

Question 8: $\frac{cos(\pi + x)\hspace{0.1cm}cos(-x)}{sin (\pi-x) \hspace{0.1cm}cos(\frac{\pi}{2}+x)}$ = cot²x

Question 9: $cos (\frac{3\pi}{2}+x)\hspace{0.1cm}cos (2\pi+x)[\hspace{0.1cm}cot (\frac{3\pi}{2}-x)+cot(2\pi+x)]$ = 1

Question 11: $cos (\frac{3\pi}{4}+x)-cos (\frac{3\pi}{4}-x)$ = – $\sqrt{2}$ sin x

Question 12: sin² 6x – sin² 4x = sin 2x sin 10x

Question 13: cos² 2x – cos² 6x = sin 4x sin 8x

Question 14: sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x