# NCRT Solutions Excercise 3

Question 1: sin2 (π/6) + cos2(π/3)-tan2(π/4)=(-1/2)

Answer 1: L.H.S.=sin2 (π/6) + cos2(π/3)-tan2(π/4)

=R.H.S

Question 2: Prove that

2sin2 (π/6) + cosec2 (7π/3)cos2(π/3)

Answer 2: L.H.S.= 2sin2 (π/6) + cosec2 (7π/3)cos2(π/3)

=R.H.S.

Question 3:

Prove that cot2 (π/6) + cosec(5π/6)+3 tan2(π/6)=6

L.H.S = cot2 (π/6) + cosec(5π/6)+3 tan2(π/6)

= 3 + 2 + 1 = 6

= R.H.S

Question 4: Prove that

= 1+1+8

= 10

= R.H.S

Question 5:

Find the value of:

(i)sin 75°

(ii) tan 15°

(i) sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

[sin (x + y) = sin x cos y + cos x sin y]

(ii)  tan 15° = tan (45° – 30°)

Question 6:  Prove that:

=sin(x+y)

= R.H.S

Question 7: Prove that:

Answer 7: It is known that

L.H.S. =

= R.H.S

Question 8: Prove that

= cot2x

= R.H.S.

Question 9:

=sin x cos x[tan x+cot x]

= 1 = R.H.S

Question 10: Prove that

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

Answer 10: L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x

= ½[2 sin(n+1)x sin(n+2)x+2 cos(n+1)x cos(n+2)x]

= (1/2) x 2 cos{(n+1)x-(n+2)x}

= cos(-x)

= cos x

= R.H.S.

Question 11: Prove that

Answer 11: It is known that

Therefore L.H.S.=

= -2 sin(3π/4)sin x

= -2 sin(π/4)sin x

= R.H.S.

Question 12: Prove that   sin26x – sin24x = sin 2x sin 10x

It is known that

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

Therefore L.H.S. = sin26x – sin24x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

= (2 sin 5x cos x) (2 cos 5x sin x)

= (2 sin 5x cos 5x) (2 sin x cos x)

= sin 10x sin 2x

= R.H.S.

Question 13: Prove that   cos22x – cos26x = sin 4x sin 8x

It is known that

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

∴ L.H.S. = cos22x – cos26x

= (cos 2x + cos 6x) (cos 2x – 6x)

= [2 cos 4x cos(-2x)][-2 sin 4x sin(-2x)]

= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]

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

= sin 8x sin 4x

= R.H.S.

Question 14:  Prove that

sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

Answer 14: L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

= 2 sin 4x cos (– 2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

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

= 2 sin 4x (2 cos2 x – 1 + 1)

= 2 sin 4x (2 cos2 x)

= 4cos2 x sin 4x

= R.H.S.

Question 15: Prove that

cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x – sin 3x)

= 2 cos 4x. cos x

L.H.S. = R.H.S.

Question 16: Prove that

It is known that

∴ L.H.S

= R.H.S.

Question 17: Prove that :

Answer 17:  It is known that

∴ L.H.S.=

= tan 4x

= R.H.S.

Question 18: Prove that

It is known that

∴ L.H.S.=

= R.H.S.

Question 19:  Prove that

It is known that

∴ L.H.S.=

= tan 2x

= R.H.S.

Question 20:  Prove that

It is known that

∴ L.H.S.=

= -2 x (-sin x)

= 2 sin x

= R.H.S.

Question 21: Prove that

L.H.S.=

= cot 3x

= R.H.S.

Question 22: Prove that

cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cot x cot 2x – (cot 2x cot x – 1) = 1

= R.H.S.

Question 23:  Prove that

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

= R.H.S.

Question 24:  Prove that:  cos 4x = 1 – 8sin2 x cos2 x

Answer 24: L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2 A]

= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]

= 1 – 8 sin2x cos2x

= R.H.S.

Question 25:  Prove that:

cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1

Answer 25: L.H.S. = cos 6x

= cos 3(2x)

= 4 cos32x – 3 cos 2x [cos 3A = 4 cos3 A – 3 cos A]

= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 x – 1]

= 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3

= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3

= 32 cos6x – 4 – 48 cos4x + 24 cos2x – 6 cos2x + 3

= 32 cos6x – 48 cos4x + 18cos2x – 1

= R.H.S.

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