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

**Prove the following trigonometric identities:**

**Question 57. tan**^{2 }A sec^{2 }B − sec^{2 }A tan^{2 }B = tan^{2 }A − tan^{2 }B

^{2 }A sec

^{2 }B − sec

^{2 }A tan

^{2 }B = tan

^{2 }A − tan

^{2 }B

**Solution:**

We have,

L.H.S. = tan

^{2}A sec^{2}B − sec^{2}A tan^{2}B= tan

^{2}A (1 + tan^{2}B) − tan^{2}B (1+ tan^{2}A)= tan

^{2 }A + tan^{2 }A tan^{2 }B − tan^{2 }B − tan^{2 }A tan^{2 }B= tan

^{2}A − tan^{2}B= 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 x**^{2} − y^{2} = a^{2 }− b^{2}.

^{2}− y

^{2}= a

^{2 }− b

^{2}.

**Solution:**

We have,

L.H.S. = x

^{2}− y^{2}= (a sec θ + b tan θ)

^{2}− (a tan θ + b sec θ)^{2}= a

^{2 }sec^{2}θ + b^{2 }tan^{2 }θ + 2ab sec θ tan θ − a^{2}tan^{2 }θ − b^{2 }sec^{2 }θ – 2ab sec θ tan θ= a

^{2 }sec^{2 }θ + b^{2 }tan^{2 }θ − a^{2 }tan^{2 }θ − b^{2}sec^{2 }θ= a

^{2 }sec^{2 }θ − b^{2 }sec^{2 }θ + b^{2 }tan^{2}θ − a^{2 }tan^{2 }θ= sec

^{2 }θ (a^{2}− b^{2}) + tan^{2 }θ (b^{2}− a^{2})= sec

^{2}θ (a^{2}− b^{2}) − tan^{2}θ (a^{2}− b^{2})= (sec

^{2 }θ − tan^{2}θ) (a^{2}− b^{2})= a

^{2}− b^{2}= 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)

= cosec

^{2 }θ − cot^{2 }θ= 1

= R.H.S.

Hence proved.

**Question 61. If T**_{n} = sin^{n} θ + cos^{n} θ, Prove that **.**

_{n}= sin

^{n}θ + cos

^{n}θ, Prove that

**Solution:**

We have,

L.H.S. =

=

=

=

=

= sin

^{2}θ cos^{2}θAnd R.H.S. =

=

=

=

=

= sin

^{2}θ cos^{2}θTherefore, 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}= tan

^{2}θ + sec^{2}θ + 2 tan θ sec θ + tan^{2}θ + sec^{2}θ – 2 tan θ sec θ= 2[tan

^{2}θ + sec^{2}θ]= 2\frac{sin^2θ}{cos^2θ}+\frac{1}{cos^2θ}

= 2\frac{1+sin^2θ}{cos^2θ}

= 2\frac{1+sin^2θ}{1-sin^2θ}

= 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))

= (sec

^{2 }A − tan^{2 }A) (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 cos

^{3}A + cos A sin^{3}A= sin A cos A (sin

^{2}A + cos^{2}A)= sin A cos A

= R.H.S.

Hence proved.

**Question 71. sec**^{4 }A (1 − sin^{4 }A) – 2 tan^{2 }A = 1

^{4 }A (1 − sin

^{4 }A) – 2 tan

^{2 }A = 1

**Solution:**

We have,

L.H.S. = sec

^{4}A (1 − sin^{4}A) – 2 tan^{2}A= sec

^{4}A – tan^{4}A – 2 tan^{4 }A= (sec

^{2}A)^{2}– tan^{4}A – 2 tan^{4}A= (1+ tan

^{2}A)^{2}− tan^{4}A − 2tan^{4}A= 1 + tan

^{4}A + 2tan^{2}A − tan^{4}A − 2tan^{4}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 x**^{2}/a^{2} + y^{2}/b^{2} = 2.

^{2}/a

^{2}+ y

^{2}/b

^{2}= 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 = a**^{3}, sec θ – cos θ = b^{3}, Prove that a^{2}b^{2} (a^{2}+ b^{2}) = 1.

^{3}, sec θ – cos θ = b

^{3}, Prove that a

^{2}b

^{2}(a

^{2}+ b

^{2}) = 1.

**Solution:**

We are given,

=> cosec θ – sin θ = a

^{3}=> 1/sin θ – sin θ = a

^{3}=> a

^{3}==> a

^{3}==> a =

On squaring both sides, we get,

=> a

^{2}=Also we have,

=> sec θ – cos θ = b

^{3}=> 1/cos θ – cos θ = b

^{3}=> b

^{3}==> b

^{3}==> b =

On squaring both sides, we get,

=> b

^{2}=So, L.H.S. = a

^{2}b^{2}(a^{2}+ b^{2})=

=

= 1

= R.H.S.

Hence proved.

**Question 76. If a cos**^{3} θ + 3a cos θ sin^{2 }θ = m and a sin^{3 }θ + 3a cos^{2 }θ sin θ = n, prove that

^{3}θ + 3a cos θ sin

^{2 }θ = m and a sin

^{3 }θ + 3a cos

^{2 }θ sin θ = n, prove that

**Solution:**

We are given,

m = a cos

^{3}θ + 3a cos θ sin^{2}θ and n = a sin^{3}θ + 3a cos^{2}θ sin θSo, L.H.S. =

= (a cos

^{3 }θ + 3a cos θ sin^{2}θ + a sin^{3}θ + 3a cos^{2}θ sin θ)^{2/3}+ (a cos^{3}θ + 3a cos θ sin^{2}θ – a sin^{3}θ – 3a cos^{2}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}[cos^{2}θ + sin^{2}θ + 2 sin θ cos θ + cos^{2 }θ + sin^{2}θ − 2 sin θ cos θ]= 2 a

^{2/3}= R.H.S.

Hence proved.

**Question 77. If x = a cos**^{3 }θ, y = b sin^{3}θ, prove that (x/a)^{2/3} + (y/b)^{2/3} = 1.

^{3 }θ, y = b sin

^{3}θ, prove that (x/a)

^{2/3}+ (y/b)

^{2/3}= 1.

**Solution:**

Given x = a cos

^{3}θ and y = b sin^{3 }θ.So, L.H.S. = (x/a)

^{2/3}+ (y/b)^{2/3}=

= (cos

^{3}θ)^{2/3}+ (sin^{3}θ)^{2/3}= cos

^{2 }θ + sin^{2}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 a**^{2} + b^{2} = m^{2} + n^{2}.

^{2}+ b

^{2}= m

^{2}+ n

^{2}.

**Solution:**

We have,

R.H.S = m

^{2}+ n^{2}= (a cos θ + b sin θ)

^{2}+ (a sin θ – b cos θ)^{2}= a

^{2}cos^{2}θ + b^{2}sin^{2}θ + 2ab sin θ cos θ + a^{2 }sin^{2 }θ + b^{2}cos^{2 }θ – 2ab sin θ cos θ= a

^{2}cos^{2 }θ + a^{2}cos^{2 }θ + b^{2}sin^{2}θ + b^{2 }cos^{2}θ= a

^{2 }(sin^{2}θ + cos^{2}θ) + b^{2 }(sin^{2}θ + cos^{2}θ)= a

^{2}+ b^{2 }= L.H.S.

Hence proved.

**Question 79. If cos A + cos**^{2 }A = 1, Prove that sin^{2 }A + sin^{4 }A = 1.

^{2 }A = 1, Prove that sin

^{2 }A + sin

^{4 }A = 1.

**Solution:**

We are given,

=> cos A + cos

^{2}A = 1=> cos A = 1 − cos

^{2 }A=> cos A = sin

^{2}A . . . . (1)Now, L.H.S. = sin

^{2}A + sin^{4}AUsing (1), we get,

= cos A + cos

^{2}A= 1

= R.H.S.

Hence proved.

**Question 80. If cos θ + cos**^{2} θ = 1, prove that sin^{12} θ + 3 sin^{10} θ + 3 sin^{8} θ + sin^{6} θ + 2 sin^{4} θ + 2 sin^{2} θ − 2 = 1.

^{2}θ = 1, prove that sin

^{12}θ + 3 sin

^{10}θ + 3 sin

^{8}θ + sin

^{6}θ + 2 sin

^{4}θ + 2 sin

^{2}θ − 2 = 1.

**Solution:**

We are given,

=> cos θ + cos

^{2}θ = 1=> cos θ = 1 − cos

^{2}θ=> cos θ = sin

^{2}θ . . . . (1)Now, L.H.S. = sin

^{12}θ + 3 sin^{10}θ + 3 sin^{8}θ + sin^{6}θ + 2 sin^{4}θ + 2 sin^{2}θ − 2= (sin

^{4}θ)^{3}+ 3 sin^{4}θ sin^{2}θ (sin^{4}θ + sin^{2}θ) + (sin^{2}θ)^{3}+ 2(sin^{2 }θ)^{2}+ 2 sin^{2}θ − 2Using (1), we get,

= (sin

^{4}θ + sin^{2}θ)^{3}+ 2cos^{2}θ + 2 cos θ − 2= ((sin

^{2}θ)^{2}+ sin^{2}θ)^{3}+ 2 cos^{2}θ + 2 cos θ – 2= (cos

^{2}θ + sin^{2}θ)^{3}+ 2 cos^{2}θ + 2 cos θ − 2= 1 + 2(cos

^{2}θ + sin^{2}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 cos

^{2}(α/2).2 cos^{2}(β/2).2 cos^{2}(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 sin**^{6} θ + cos^{6} θ **= ****.**

^{6}θ + cos

^{6}θ

**Solution:**

We are given,

=> sin θ + cos θ = x

On squaring both sides, we get,

=> (sin θ + cos θ)

^{2}= x^{2}=> sin

^{2 }θ + cos^{2 }θ + 2 sin θ cos θ = x^{2}=> 2 sin θ cos θ = x

^{2}− 1=> sin θ cos θ = (x

^{2}− 1)/2 . . . . (1)We know,

=> sin

^{2}θ + cos^{2}θ = 1Cubing on both sides, we get

=> (sin

^{2}θ + cos^{2}θ)^{3}= 1^{3}=> sin

^{6}θ + cos^{6}θ + 3 sin^{2}θ cos^{2}θ (sin^{2}θ + cos^{2}θ) = 1=> sin

^{6}θ + cos^{6}θ = 1 – 3 sin^{2}θ cos^{2}θFrom (1), we get,

=> sin

^{6}θ + cos^{6}θ = 1 –=> sin

^{6}θ + cos^{6}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, x**^{2}/a^{2} + y^{2}/b^{2} − z^{2}/c^{2} = 1.

^{2}/a

^{2}+ y

^{2}/b

^{2}− z

^{2}/c

^{2}= 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,

x

^{2 }= a^{2 }sec^{2 }θ cos^{2}ϕ or x^{2}/a^{2}= sec^{2 }θ cos^{2}ϕ . . . . (1)y

^{2}= b^{2}sec^{2 }θ sin^{2}ϕ or y^{2}/b^{2}= sec^{2 }θ sin^{2}ϕ . . . . (2)z

^{2}= c^{2}tan^{2}ϕ or z^{2}/c^{2}= tan^{2}ϕ . . . . (3)Now L.H.S. = x

^{2}/a^{2}+ y^{2}/b^{2}− z^{2}/c^{2}Using (1), (2) and (3), we get,

= sec

^{2}θ cos^{2}ϕ + sec^{2}θ sin^{2}ϕ − tan^{2}ϕ= sec

^{2}θ (cos^{2}ϕ + sin^{2}ϕ) − tan^{2}ϕ= sec

^{2}θ (1) − tan^{2}ϕ= sec

^{2 }θ − tan^{2 }θ= 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=> sin

^{2 }θ + 4 cos^{2 }θ + 4 sin θ cos θ = 1=> 4 cos

^{2}θ + 4 sin θ cos θ = 1 – sin^{2 }θ=> 4 cos

^{2 }θ + 4 sin θ cos θ – cos^{2 }θ = 0=> 3 cos

^{2 }θ + 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 sin

^{2 }θ + cos^{2 }θ – 4 sin θ cos θFrom (1), we get,

= 4 sin

^{2 }θ + cos^{2 }θ + 3 cos^{2}θ= 4 sin

^{2}θ + 4 cos^{2}θ= 4(sin

^{2 }θ + cos^{2 }θ)= 4

So, we have,

=> (2 sin θ – cos θ)

^{2}= 4=> 2 sin θ – cos θ = 2

Hence proved.

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