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# NCRT Solutions Excercise 3

**Question 1:** sin^{2} (π/6) + cos^{2}(π/3)-tan^{2}(π/4)=(-1/2)

**Answer 1:** L.H.S.=sin^{2 }(π/6) + cos^{2}(π/3)-tan^{2}(π/4)

=R.H.S

**Question 2:** Prove that

2sin^{2 }(π/6) + cosec^{2 }(7π/3)cos^{2}(π/3)

**Answer 2:** L.H.S.= 2sin^{2 }(π/6) + cosec^{2 }(7π/3)cos^{2}(π/3)

=R.H.S.

**Question 3: **

Prove that cot^{2 }(π/6) + cosec(5π/6)+3 tan^{2}(π/6)=6

**Answer 3:**

L.H.S = cot^{2 }(π/6) + cosec(5π/6)+3 tan^{2}(π/6)

= 3 + 2 + 1 = 6

= R.H.S

**Question 4: **Prove that

**Answer 4:** L.H.S =

= 1+1+8

= 10

= R.H.S

**Question 5:**

Find the value of:

(i)sin 75°

(ii) tan 15°

**Answer 5:**

(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:

**Answer 6:**

=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

**Answer 8:** L.H.S =

= cot^{2}x

= R.H.S.

**Question 9:**

**Answer 9:** L.H.S =

=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 sin^{2}6x – sin^{2}4x = sin 2x sin 10x

**Answer 12:**

It is known that

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

Therefore L.H.S. = sin^{2}6x – sin^{2}4x

= (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 cos^{2}2x – cos^{2}6x = sin 4x sin 8x

**Answer 13:**

It is known that

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

∴ L.H.S. = cos^{2}2x – cos^{2}6x

= (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 = 4cos^{2} 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 cos^{2} x – 1 + 1)

= 2 sin 4x (2 cos^{2} x)

= 4cos^{2} x sin 4x

= R.H.S.

**Question 15:** Prove that

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

**Answer 15: **

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

**Answer 16:**

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

**Answer 18:**

It is known that

∴ L.H.S.=

= R.H.S.

**Question 19:** Prove that

**Answer 19:**

It is known that

∴ L.H.S.=

= tan 2x

= R.H.S.

**Question 20:** Prove that

**Answer 20:**

It is known that

∴ L.H.S.=

= -2 x (-sin x)

= 2 sin x

= R.H.S.

**Question 21:** Prove that

**Answer 21:**

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

**Answer 22:**

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

**Answer 23:**

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

= R.H.S.

**Question 24:** Prove that: cos 4x = 1 – 8sin^{2} x cos^{2} x

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

= cos 2(2x)

= 1 – 2 sin^{2} 2x [cos 2A = 1 – 2 sin^{2} A]

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

= 1 – 8 sin^{2}x cos^{2}x

= R.H.S.

**Question 25:** Prove that:

cos 6x = 32 cos^{6} x – 48 cos^{4} x + 18 cos^{2} x – 1

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

= cos 3(2x)

= 4 cos^{3}2x – 3 cos 2x [cos 3A = 4 cos^{3} A – 3 cos A]

= 4 [(2 cos^{2} x – 1)^{3} – 3 (2 cos^{2} x – 1) [cos 2x = 2 cos^{2} x – 1]

= 4 [(2 cos^{2} x)^{3} – (1)^{3} – 3 (2 cos^{2} x)^{2} + 3 (2 cos^{2} x)] – 6cos^{2} x + 3

= 4 [8cos^{6}x – 1 – 12 cos^{4}x + 6 cos^{2}x] – 6 cos^{2}x + 3

= 32 cos^{6}x – 4 – 48 cos^{4}x + 24 cos^{2}x – 6 cos^{2}x + 3

= 32 cos^{6}x – 48 cos^{4}x + 18cos^{2}x – 1

= R.H.S.