NCERT/CBSE
- NCERT Solutions For Class 11

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Learn Exercise 3.3 with Free Lessons & Tips

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Question 1:

Prove that

Solution 1:

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

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Charlie

Question 2:

Solution 2:

L.H.S. =

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Question 3:

Prove that

Solution 3:

L.H.S. =

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Question 4:

Prove that

Solution 4:

L.H.S. =

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Question 5:

Prove that

Solution 5:

L.H.S =

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Question 6:

Find the value of:

(i) sin 75°

(ii) tan 15°

Solution 6:

(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°)

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Swapna Shree

Question 7:

Prove that:

Solution 7:

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Question 8:

Prove that:

Solution 8:

It is known that

∴L.H.S. =

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Question 9:

Solution 9:

L.H.S. =

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Swapna Shree

Question 10:

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

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

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Swapna Shree

Question 11:
Prove that

Solution 11:
It is known that.

∴L.H.S. =

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

Solution 12:
It is known that

∴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.

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

Solution 13:
It is known that

∴L.H.S. = cos² 2x – cos² 6x

= (cos 2x + cos 6x) (cos 2x – 6x)

= [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.

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Question 14:
Prove that sin 2x + 2sin 4x + sin 6x = 4cos²x sin 4x

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

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

= 2 sin 4x cos (â€“ 2x) + 2 sin 4x

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

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

= 2 sin 4x (2 cos²x â€“ 1 + 1)

= 2 sin 4x (2 cos²x)

= 4cos²x sin 4x

= R.H.S.

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Question 15:
Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution 15:
L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x â€“ sin 3x)

= 2 cos 4x. cos x

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

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Question 16:
Prove that

Solution 16:
It is known that

∴L.H.S =

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Question 17:
Prove that

Solution 17:
It is known that

∴L.H.S. =

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Swapna Shree

Question 18:
Prove that

Solution 18:
It is known that

∴L.H.S. =

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Swapna Shree

Question 19:
Prove that

Solution 19:
It is known that

∴L.H.S. =

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Swapna Shree

Question 20:
Prove that

Solution 20:
It is known that

∴L.H.S. =

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Question 21:
Prove that

Solution 21:
L.H.S. =

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Question 22:
Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution 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.

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Question 23:
Prove that

Solution 23:
It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

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Question 24:
Prove that cos 4x = 1 – 8sin²x cos²x

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

= cos 2(2x)

= 1 – 2 sin² 2x [cos 2A = 1 – 2 sin²A]

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

= 1 – 8 sin²x cos²x

= R.H.S.

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Question 25:

Prove that: cos 6x = 32 cos⁶x – 48 cos⁴x + 18 cos²x – 1

Solution 25:

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos³ 2x – 3 cos2x [cos 3A = 4 cos³A – 3 cosA]

= 4 [(2 cos²x – 1)³ – 3 (2 cos²x – 1) [cos 2x = 2 cos²x – 1]

= 4 [(2 cos²x)³ – (1)³ – 3 (2 cos²x)² + 3 (2 cos²x)] – 6cos²x + 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 + 18 cos²x – 1

= R.H.S.

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Swapna Shree

All Exercises - Chapter 3 -Trigonometric Functions

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Miscellaneous Exercise 3

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Learn Exercise 3.3 with Free Lessons & Tips

All Exercises - Chapter 3 -Trigonometric Functions

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