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Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.3 | Set 1
  • Last Updated : 16 Apr, 2021

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 cos2 x)

= 4 sin (4x) cos2 x

Hence, LHS = RHS

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