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

• Last Updated : 16 Apr, 2021

### Question 1: sin2+cos2– tan2=

Solution:

Taking LHS in consideration, we get

= sin2+cos2– tan2

Substituting the values,

=

=– 1

=

Hence, LHS = RHS

### Question 2: 2sin2+cosec2cos2=

Solution:

Taking LHS in consideration, we get

= 2sin2+cosec2cos2

= 2sin2+ (- coseccos2

Substituting the values,

=

= 2+ 4

=+ 1

=

Hence, LHS = RHS

### Question 3: cot2+ cosec+ 3tan2= 6

Solution:

Taking LHS in consideration, we get

= cot2+ cosec+ 3tan2

= cot2+ cosec+ 3tan2

Substituting the values,

= (√3)2 + 2 + 3

= 3 + 2 + 3

= 6

Hence, LHS = RHS

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

Solution:

Taking LHS in consideration, we get

= 2sin2+ 2cos2+ 2sec2

= 2sin2+ 2cos2+ 2sec2

Substituting the values,

=

= 2+ 2(4)

= 1 + 1 + 8

= 10

Hence, LHS = RHS

### (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°) =

sin(75°) =

sin(75°) =

### (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) =

tan(45° – 30°) =

Substituting values, we get

tan(15°) =

tan(15°) =

tan(15°) =

Now rationalizing the denominator, multiply and divide by

tan(15°) =

tan(15°) =

tan(15°) =

tan(15°) =

tan(15°) =

tan(15°) = 2 –

### Question 6:= sin (x+y)

Solution:

Taking LHS in consideration, we get

=

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

=

=

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

=

= cos

= cos

= sin (x+y) (As cos= sin θ)

Hence, LHS = RHS

### Question 7:

Solution:

Taking LHS in consideration, we get

As, using Factorisation Formulae of tan, we have

tan (A+B) =and,

tan (A-B) =

Now, substituting the values

=

=

=

=

Hence, LHS = RHS

### Question 8:= cot2x

Solution:

Taking LHS in consideration, we get

=

As, we know these standard values

cos(-x) = cos x

cos= – cos x

sin= sin x

cos= – sin x

Substituting these values, we have

=

=

= cot2 x

Hence, LHS = RHS

### Question 9:= 1

Solution:

Taking LHS in consideration, we get

=

As, we know these standard values

cos= sin x

cos= cos x

cot= cot x

cot= tan x

Substituting the values, we have

=

=

=

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

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

=(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

=(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:= –sin x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

cos A – cos B = 2 sinsin

Substituting the values,

=

= 2 sinsin

= 2 sinsin (-x)

= 2 sinsin (-x)

= 2 ( sin) sin (-x)

= 2(- sin x)

=

Rationalizing the denominator, by multiplying and dividing by

=

=

=

=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

=(sin 6x sin 6x – sin 4x sin 4x)

=(2 sin 6x sin 6x – 2 sin 4x sin 4x)

Using the identity, we can simplify

=[(cos(6x-6x) – cos(6x+6x)) – (cos(4x-4x) – cos(4x+4x))]

=[(cos(0) – cos(12x)) – (cos(0) – cos(8x))]

=[1 – cos(12x) – 1 + cos(8x)] (As, cos 0 = 1)

=[cos(8x) – cos (12x)]

Now, using the identity

cos A – cos B = 2 sinsin

Substituting the values, we have

=

= sinsin

= 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

=(cos 2x cos 2x – cos 6x cos 6x)

=(2 cos 2x cos 2x – 2 cos 6x cos 6x)

Using the identity, we can simplify

=[(cos(2x+2x) + cos(2x-2x)) – (cos(6x+6x) + cos(6x-6x))]

=[(cos(2x+2x) + cos(0)) – (cos(6x+6x) + cos(0))]

=[(cos(4x) + 1 – cos(12x) – 1)] (As, cos 0 = 1)

=[cos(4x) – cos(12x)]

Now, using the identity

cos A – cos B = 2 sinsin

Substituting the values, we have

=

= sinsin

= 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 sincos

= 2 sincos+ 2 sin 4x

= 2 sincos+ 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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