Class 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.1 | Set 3

Last Updated : 23 Jul, 2025

The Trigonometric identities are fundamental formulas in trigonometry that relate the angles and sides of the triangles. These identities simplify trigonometric expressions and solve the equations involving trigonometric functions. In Class 10 RD Sharma's Chapter 6 on the Trigonometric Identities Exercise 6.1 | Set 3 provides practice problems to the reinforce these concepts and apply them effectively.

Trigonometric Identities

The Trigonometric identities are equations involving the trigonometric functions that hold for all angles. Here are four key trigonometric identities:

Pythagorean Identity:

This identity arises from the Pythagorean theorem applied to the unit circle.

Reciprocal Identities:

These identities express the reciprocal relationships between the trigonometric functions.

Quotient Identities:

These identities define the tangent and cotangent functions in terms of the sine and cosine.

Co-Function Identities:

These identities relate the sine and cosine of the complementary angles.

Prove the following trigonometric identities:

Question 57. tan²A sec²B − sec²A tan²B = tan²A − tan²B

Solution:

We have,
L.H.S. = tan² A sec² B − sec² A tan² B
= tan² A (1 + tan² B) − tan² B (1+ tan² A)
= tan²A + tan²A tan²B − tan²B − tan²A tan²B
= tan²A − tan²B
= R.H.S.
Hence proved.

Question 58. If x = a sec θ + b tan θ and y = a tan θ + b sec θ, Prove that x² − y² = a²− b².

Solution:

We have,
L.H.S. = x² − y²
= (a sec θ + b tan θ)² − (a tan θ + b sec θ)²
= a²sec² θ + b²tan²θ + 2ab sec θ tan θ − a² tan²θ − b²sec²θ – 2ab sec θ tan θ
= a²sec²θ + b²tan²θ − a²tan²θ − b² sec²θ
= a²sec²θ − b²sec²θ + b²tan²θ − a²tan²θ
= sec²θ (a² − b²) + tan²θ (b² − a²)
= sec²θ (a² − b²) − tan²θ (a² − b²)
= (sec²θ − tan²θ) (a² − b²)
= a² − b²
= R.H.S.
Hence proved.

Question 59. If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = 3.

Solution:

We are given,
=> 3 sin θ + 5 cos θ = 5
=> 3 sin θ = 5 (1 − cos θ)
=> 3 sin θ =
=> 3 sin θ =
=> 3 sin θ =
=> 3 (1 + cos θ) = 5 sin θ
=> 3 + 3 cos θ = 5 sin θ
=> 5 sin θ − 3 cos θ = 3
Hence proved.

Question 60. If cosec θ + cot θ = m and cosec θ – cot θ = n, prove that m n = 1.

Solution:

We have,
L.H.S. = m n
= (cosec θ + cot θ) (cosec θ – cot θ)
= cosec²θ − cot²θ
= 1
= R.H.S.
Hence proved.

Question 61. If T_n = sinⁿ θ + cosⁿ θ, Prove that .

Solution:

We have,
L.H.S. =
=
=
=
=
= sin² θ cos² θ
And R.H.S. =
=
=
=
=
= sin² θ cos² θ
Therefore, L.H.S. = R.H.S.
Hence proved.

Question 62.

Solution:

We have,
L.H.S. =
= (tan θ + sec θ)² + (tan θ – sec θ)²
= tan²θ + sec²θ + 2 tan θ sec θ + tan² θ + sec²θ – 2 tan θ sec θ
= 2[tan² θ + sec² θ]
= 2[]
= 2[]
= 2[]
= R.H.S.
Hence proved.

Question 63.

Solution:

We have,
L.H.S. =
=
=
=
=
=
=
=
=
= R.H.S.
Hence proved.

Question 64. (i)

Solution:

We have,
L.H.S. =
=
=
=
=
=
=
=
=
=
=
=
= R.H.S.
Hence proved.

(ii)

Solution:

We have,
L.H.S. =
=
=
=
= sec θ − tan θ
= 1/cos θ − sin θ/cos θ
=
= R.H.S.
Hence proved.

Question 65. (sec A + tan A − 1) (sec A - tan A + 1) = 2 tan A

Solution:

We have,
L.H.S. = (sec A + tan A − 1) (sec A - tan A + 1)
= [sec A + tan A − (sec A + tan A) (sec A – tan A)] [sec A – tan A + (sec A – tan A)(sec A + tan A)]
= (sec A + tan A) (1 − (sec A – tan A)) (sec A – tan A) (1 + (sec A + tan A))
= (sec²A − tan²A) (1 – sec A + tan A) (1 + sec A + tan A)
= (1 – sec A + tan A) (1 + sec A + tan A)
= (1 – 1/cos A + sin A/cos A) (1 + 1/cos A + sin A/cos A)
=
=
=
=
= sin A/cos A
= 2 tan A
= R.H.S.
Hence proved.

Question 66. (1 + cot A − cosec A)(1 + tan A + sec A) = 2

Solution:

We have,
L.H.S. = (1 + cot A − cosec A)(1 + tan A + sec A)
= (1 + cos A/sin A − 1/sin A)(1 + sin A/cos A + 1/cos A)
=
=
=
=
= 2
= R.H.S.
Hence proved.

Question 67. (cosec θ – sec θ) (cot θ – tan θ) = (cosec θ + sec θ) (sec θ cosec θ − 2)

Solution:

We have,
L.H.S. = (cosec θ – sec θ) (cot θ – tan θ)
=
=
=
And R.H.S. = (cosec θ + sec θ) (sec θ cosec θ − 2)
=
=
=
=
=
Therefore, L.H.S. = R.H.S.
Hence proved.

Question 68.

Solution:

We have,
L.H.S. =
=
=
=
= cosec A − sec A
= R.H.S.
Hence proved.

Question 69.

Solution:

We have,
L.H.S. =
=
=
=
=
=
=
= 1
= R.H.S.
Hence proved.

Question 70.

Solution:

We have,
=
=
=
= sin A cos³ A + cos A sin³ A
= sin A cos A (sin² A + cos² A)
= sin A cos A
= R.H.S.
Hence proved.

Question 71. sec⁴A (1 − sin⁴A) – 2 tan²A = 1

Solution:

We have,
L.H.S. = sec⁴ A (1 − sin⁴ A) – 2 tan² A
= sec⁴ A - tan⁴ A – 2 tan⁴A
= (sec² A)² - tan⁴ A – 2 tan⁴ A
= (1+ tan² A)² − tan⁴ A − 2tan⁴ A
= 1 + tan⁴ A + 2tan² A − tan⁴ A − 2tan⁴ A
= 1
= R.H.S.
Hence proved.

Question 72.

Solution:

We have,
L.H.S. =
=
=
=
And R.H.S. =
=
=
=
=
=
=
=
Therefore, L.H.S. = R.H.S.
Hence proved.

Question 73. (1 + cot A + tan A) (sin A – cos A) = sin A tan A – cos A cot A

Solution:

We have,
L.H.S. = (1 + cot A + tan A) (sin A – cos A)
= sin A – cos A + cot A sin A – cot A cos A + sin A tan A – tan A cos A
= sin A – cos A + cos A – cot A cos A + sin A tan A – sin A
= sin A tan A – cos A cot A
= R.H.S
Hence proved.

Question 74. If x cos θ/a + y sin θ/b = 1 and x cos θ/a – y sin θ/b = 1, then prove that x²/a² + y²/b² = 2.

Solution:

We have,
x cos θ/a + y sin θ/b = 1 . . . . (1)
x cos θ/a – y sin θ/b = 1 . . . . (2)
On squaring both sides of (1) and (2) and adding them we get,
=> (x cos θ/a + y sin θ/b)² + (x cos θ/a – y sin θ/b)² = 1 + 1
=>  = 2
=>  = 2
=>  = 2
Hence proved.

Question 75. If cosec θ – sin θ = a³, sec θ – cos θ = b³, Prove that a²b² (a²+ b²) = 1.

Solution:

We are given,
=> cosec θ – sin θ = a³
=> 1/sin θ – sin θ = a³
=> a³ =
=> a³ =
=> a =
On squaring both sides, we get,
=> a² =
Also we have,
=> sec θ – cos θ = b³
=> 1/cos θ – cos θ = b³
=> b³ =
=> b³ =
=> b =
On squaring both sides, we get,
=> b² =
So, L.H.S. = a²b² (a²+ b²)
=
=
= 1
= R.H.S.
Hence proved.

Question 76. If a cos³ θ + 3a cos θ sin²θ = m and a sin³θ + 3a cos²θ sin θ = n, prove that

Solution:

We are given,
m = a cos³ θ + 3a cos θ sin² θ and n = a sin³ θ + 3a cos² θ sin θ
So, L.H.S. =
= (a cos³θ + 3a cos θ sin² θ + a sin³ θ + 3a cos² θ sin θ)^2/3 + (a cos³ θ + 3a cos θ sin² θ – a sin³ θ – 3a cos² θ sin θ)^2/3
= a^2/3 ((cos θ + sin θ)³)^2/3+ a^2/3 ((cos θ − sin θ)³)^2/3
= a^2/3 [(cos θ + sin θ)² + (cos θ − sin θ)²]
= a^2/3 [cos² θ + sin² θ + 2 sin θ cos θ + cos²θ + sin² θ − 2 sin θ cos θ]
= 2 a^2/3
= R.H.S.
Hence proved.

Question 77. If x = a cos³θ, y = b sin³θ, prove that (x/a)^2/3 + (y/b)^2/3 = 1.

Solution:

Given x = a cos³ θ and y = b sin³θ.
So, L.H.S. = (x/a)^2/3 + (y/b)^2/3
=
= (cos³ θ)^2/3 + (sin³ θ)^2/3
= cos²θ + sin² θ
= 1
= R.H.S.
Hence proved.

Question 78. If a cos θ + b sin θ = m and a sin θ – b cos θ = n, Prove that a² + b² = m² + n².

Solution:

We have,
R.H.S = m² + n²
= (a cos θ + b sin θ)² + (a sin θ – b cos θ)²
= a² cos² θ + b² sin²θ + 2ab sin θ cos θ + a²sin²θ + b² cos²θ – 2ab sin θ cos θ
= a² cos²θ + a² cos²θ + b² sin² θ + b²cos² θ
= a²(sin² θ + cos² θ) + b²(sin² θ + cos² θ)
= a² + b²
= L.H.S.
Hence proved.

Question 79. If cos A + cos²A = 1, Prove that sin²A + sin⁴A = 1.

Solution:

We are given,
=> cos A + cos² A = 1
=> cos A = 1 − cos²A
=> cos A = sin² A . . . . (1)
Now, L.H.S. = sin² A + sin⁴ A
Using (1), we get,
= cos A + cos² A
= 1
= R.H.S.
Hence proved.

Question 80. If cos θ + cos² θ = 1, prove that sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2 = 1.

Solution:

We are given,
=> cos θ + cos² θ = 1
=> cos θ = 1 − cos² θ
=> cos θ = sin² θ . . . . (1)
Now, L.H.S. = sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2
= (sin⁴ θ)³ + 3 sin⁴ θ sin² θ (sin⁴ θ + sin² θ) + (sin² θ)³ + 2(sin²θ)² + 2 sin² θ − 2
Using (1), we get,
= (sin⁴ θ + sin² θ)³ + 2cos² θ + 2 cos θ − 2
= ((sin² θ)² + sin² θ)³ + 2 cos² θ + 2 cos θ – 2
= (cos² θ + sin² θ)³ + 2 cos² θ + 2 cos θ − 2
= 1 + 2(cos² θ + sin² θ) − 2
= 1 + 2(1) −2
= 1
= R.H.S.
Hence proved.

Question 81. Given that: (1 + cos α)(1 + cos β)(1 + cos γ) = (1 – cos α)(1 – cos β)(1 – cos γ). Show that one of the values of each member of this equality is sin α sin β sin γ.

Solution:

We have,
= (1 + cos α)(1 + cos β)(1 + cos γ)
= 2 cos² (α/2).2 cos² (β/2).2 cos² (γ/2)
=
=
=
Therefore, sin α sin β sin γ is the member of equality.
Hence proved.

Question 82. If sin θ + cos θ = x, prove that sin⁶ θ + cos⁶ θ= .

Solution:

We are given,
=> sin θ + cos θ = x
On squaring both sides, we get,
=> (sin θ + cos θ)² = x²
=> sin²θ + cos²θ + 2 sin θ cos θ = x²
=> 2 sin θ cos θ = x² − 1
=> sin θ cos θ = (x² − 1)/2 . . . . (1)
We know,
=> sin² θ + cos² θ = 1
Cubing on both sides, we get
=> (sin² θ + cos² θ)³ = 1³
=> sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ (sin² θ + cos² θ) = 1
=> sin⁶ θ + cos⁶ θ = 1 – 3 sin² θ cos²θ
From (1), we get,
=> sin⁶ θ + cos⁶ θ = 1 –
=> sin⁶ θ + cos⁶ θ =
Hence proved.

Question 83. If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan ϕ, show that, x²/a² + y²/b² − z²/c² = 1.

Solution:

We are given, x = a sec θ cos ϕ, y = b sec θ sin ϕ, z = c tan ϕ
On squaring x, y, z, we get,
x²= a²sec²θ cos² ϕ or x²/a² = sec²θ cos² ϕ . . . . (1)
y² = b² sec²θ sin² ϕ or y²/b² = sec²θ sin² ϕ . . . . (2)
z² = c² tan² ϕ or z²/c² = tan² ϕ . . . . (3)
Now L.H.S. = x²/a² + y²/b² − z²/c²
Using (1), (2) and (3), we get,
= sec² θ cos² ϕ + sec² θ sin² ϕ − tan² ϕ
= sec²θ (cos² ϕ + sin² ϕ) − tan² ϕ
= sec²θ (1) − tan² ϕ
= sec²θ − tan²θ
= 1
= R.H.S.
Hence proved.

Question 84. If sin θ + 2 cos θ. Prove that 2 sin θ – cos θ = 2.

Solution:

We are given, sin θ + 2 cos θ = 1
On squaring both sides, we get,
=> (sin θ + 2 cos θ)² = 12
=> sin²θ + 4 cos²θ + 4 sin θ cos θ = 1
=> 4 cos² θ + 4 sin θ cos θ = 1 – sin²θ
=> 4 cos²θ + 4 sin θ cos θ – cos²θ = 0
=> 3 cos²θ + 4 sin θ cos θ = 0 . . . . (1)
We have, L.H.S. = 2 sin θ – cos θ
On squaring L.H.S., we get,
= (2 sin θ – cos θ)²
= 4 sin²θ + cos²θ – 4 sin θ cos θ
From (1), we get,
= 4 sin²θ + cos²θ + 3 cos²θ
= 4 sin² θ + 4 cos² θ
= 4(sin²θ + cos²θ)
= 4
So, we have,
=> (2 sin θ – cos θ)² = 4
=> 2 sin θ – cos θ = 2
Hence proved.

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