# 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 âˆ’ sec^{2 }A tan^{2 }B = tan^{2 }A âˆ’ tan^{2 }B

^{2 }A sec

^{2 }B âˆ’ sec

^{2 }A tan

^{2 }B = tan

^{2 }A âˆ’ tan

^{2 }B

**Solution:**

We have,

L.H.S. = tan

^{2}A sec^{2}B âˆ’ sec^{2}A tan^{2}B= tan

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

^{2 }A + tan^{2 }A tan^{2 }B âˆ’ tan^{2 }B âˆ’ tan^{2 }A tan^{2 }B= tan

^{2}A âˆ’ tan^{2}B= R.H.S.

Hence proved.

**Question 58. If x = a sec Î¸ + b tan Î¸ and y = a tan Î¸ + b sec Î¸, Prove that x**^{2} âˆ’ y^{2} = a^{2 }âˆ’ b^{2}.

^{2}âˆ’ y

^{2}= a

^{2 }âˆ’ b

^{2}.

**Solution:**

We have,

L.H.S. = x

^{2}âˆ’ y^{2}= (a sec Î¸ + b tan Î¸)

^{2}âˆ’ (a tan Î¸ + b sec Î¸)^{2}= a

^{2 }sec^{2}Î¸ + b^{2 }tan^{2 }Î¸ + 2ab sec Î¸ tan Î¸ âˆ’ a^{2}tan^{2 }Î¸ âˆ’ b^{2 }sec^{2 }Î¸ â€“ 2ab sec Î¸ tan Î¸= a

^{2 }sec^{2 }Î¸ + b^{2 }tan^{2 }Î¸ âˆ’ a^{2 }tan^{2 }Î¸ âˆ’ b^{2}sec^{2 }Î¸= a

^{2 }sec^{2 }Î¸ âˆ’ b^{2 }sec^{2 }Î¸ + b^{2 }tan^{2}Î¸ âˆ’ a^{2 }tan^{2 }Î¸= sec

^{2 }Î¸ (a^{2}âˆ’ b^{2}) + tan^{2 }Î¸ (b^{2}âˆ’ a^{2})= sec

^{2}Î¸ (a^{2}âˆ’ b^{2}) âˆ’ tan^{2}Î¸ (a^{2}âˆ’ b^{2})= (sec

^{2 }Î¸ âˆ’ tan^{2}Î¸) (a^{2}âˆ’ b^{2})= a

^{2}âˆ’ b^{2}= R.H.S.

Hence proved.

**Question 59. If 3 sin Î¸ + 5 cos Î¸ = 5, prove that 5 sin Î¸ â€“ 3 cos Î¸ = 3.**

**Solution:**

We are given,

=> 3 sin Î¸ + 5 cos Î¸ = 5

=> 3 sin Î¸ = 5 (1 âˆ’ cos Î¸)

=> 3 sin Î¸ =

=> 3 sin Î¸ =

=> 3 sin Î¸ =

=> 3 (1 + cos Î¸) = 5 sin Î¸

=> 3 + 3 cos Î¸ = 5 sin Î¸

=> 5 sin Î¸ âˆ’ 3 cos Î¸ = 3

Hence proved.

**Question 60. If cosec Î¸ + cot Î¸ = m and cosec Î¸ â€“ cot Î¸ = n, prove that m n = 1. **

**Solution:**

We have,

L.H.S. = m n

= (cosec Î¸ + cot Î¸) (cosec Î¸ â€“ cot Î¸)

= cosec

^{2 }Î¸ âˆ’ cot^{2 }Î¸= 1

= R.H.S.

Hence proved.

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

_{n}= sin

^{n}Î¸ + cos

^{n}Î¸, Prove that

**Solution:**

We have,

L.H.S. =

=

=

=

=

= sin

^{2}Î¸ cos^{2}Î¸And R.H.S. =

=

=

=

=

= sin

^{2}Î¸ cos^{2}Î¸Therefore, L.H.S. = R.H.S.

Hence proved.

**Question 62. **

**Solution:**

We have,

L.H.S. =

= (tan Î¸ + sec Î¸)

^{2}+ (tan Î¸ â€“ sec Î¸)^{2}= tan

^{2}Î¸ + sec^{2}Î¸ + 2 tan Î¸ sec Î¸ + tan^{2}Î¸ + sec^{2}Î¸ â€“ 2 tan Î¸ sec Î¸= 2[tan

^{2}Î¸ + sec^{2}Î¸]= 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 Î¸ âˆ’ tan Î¸

= 1/cos Î¸ âˆ’ sin Î¸/cos Î¸

=

= R.H.S.

Hence proved.

**Question 65. (sec A + tan A âˆ’ 1) (sec A – tan A + 1) = 2 tan A **

**Solution:**

We have,

L.H.S. = (sec A + tan A âˆ’ 1) (sec A – tan A + 1)

= [sec A + tan A âˆ’ (sec A + tan A) (sec A â€“ tan A)] [sec A â€“ tan A + (sec A â€“ tan A)(sec A + tan A)]

= (sec A + tan A) (1 âˆ’ (sec A â€“ tan A)) (sec A â€“ tan A) (1 + (sec A + tan A))

= (sec

^{2 }A âˆ’ tan^{2 }A) (1 â€“ sec A + tan A) (1 + sec A + tan A)= (1 â€“ sec A + tan A) (1 + sec A + tan A)

= (1 â€“ 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 âˆ’ cosec A)(1 + tan A + sec A) = 2 **

**Solution:**

We have,

L.H.S. = (1 + cot A âˆ’ cosec A)(1 + tan A + sec A)

= (1 + cos A/sin A âˆ’ 1/sin A)(1 + sin A/cos A + 1/cos A)

=

=

=

=

= 2

= R.H.S.

Hence proved.

**Question 67. (cosec Î¸ â€“ sec Î¸) (cot Î¸ â€“ tan Î¸) = (cosec Î¸ + sec Î¸) (sec Î¸ cosec Î¸ âˆ’ 2) **

**Solution:**

We have,

L.H.S. = (cosec Î¸ â€“ sec Î¸) (cot Î¸ â€“ tan Î¸)

=

=

=

And R.H.S. = (cosec Î¸ + sec Î¸) (sec Î¸ cosec Î¸ âˆ’ 2)

=

=

=

=

=

Therefore, L.H.S. = R.H.S.

Hence proved.

**Question 68. **

**Solution:**

We have,

L.H.S. =

=

=

=

= cosec A âˆ’ 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 âˆ’ sin^{4 }A) â€“ 2 tan^{2 }A = 1

^{4 }A (1 âˆ’ sin

^{4 }A) â€“ 2 tan

^{2 }A = 1

**Solution:**

We have,

L.H.S. = sec

^{4}A (1 âˆ’ sin^{4}A) â€“ 2 tan^{2}A= sec

^{4}A – tan^{4}A â€“ 2 tan^{4 }A= (sec

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

^{2}A)^{2}âˆ’ tan^{4}A âˆ’ 2tan^{4}A= 1 + tan

^{4}A + 2tan^{2}A âˆ’ tan^{4}A âˆ’ 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 â€“ cos A) = sin A tan A â€“ cos A cot A**

**Solution:**

We have,

L.H.S. = (1 + cot A + tan A) (sin A â€“ cos A)

= sin A â€“ cos A + cot A sin A â€“ cot A cos A + sin A tan A â€“ tan A cos A

= sin A â€“ cos A + cos A â€“ cot A cos A + sin A tan A â€“ sin A

= sin A tan A â€“ cos A cot A

= R.H.S

Hence proved.

**Question 74. If x cos Î¸/a + y sin Î¸/b = 1 and x cos Î¸/a â€“ y sin Î¸/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 Î¸/a + y sin Î¸/b = 1 . . . . (1)

x cos Î¸/a â€“ y sin Î¸/b = 1 . . . . (2)

On squaring both sides of (1) and (2) and adding them we get,

=> (x cos Î¸/a + y sin Î¸/b)

^{2}+ (x cos Î¸/a â€“ y sin Î¸/b)^{2}= 1 + 1=> = 2

=> = 2

=> = 2

Hence proved.

**Question 75. If cosec Î¸ â€“ sin Î¸ = a**^{3}, sec Î¸ â€“ cos Î¸ = b^{3}, Prove that a^{2}b^{2} (a^{2}+ b^{2}) = 1.

^{3}, sec Î¸ â€“ cos Î¸ = b

^{3}, Prove that a

^{2}b

^{2}(a

^{2}+ b

^{2}) = 1.

**Solution:**

We are given,

=> cosec Î¸ â€“ sin Î¸ = a

^{3}=> 1/sin Î¸ â€“ sin Î¸ = a

^{3}=> a

^{3}==> a

^{3}==> a =

On squaring both sides, we get,

=> a

^{2}=Also we have,

=> sec Î¸ â€“ cos Î¸ = b

^{3}=> 1/cos Î¸ â€“ cos Î¸ = 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} Î¸ + 3a cos Î¸ sin^{2 }Î¸ = m and a sin^{3 }Î¸ + 3a cos^{2 }Î¸ sin Î¸ = n, prove that

^{3}Î¸ + 3a cos Î¸ sin

^{2 }Î¸ = m and a sin

^{3 }Î¸ + 3a cos

^{2 }Î¸ sin Î¸ = n, prove that

**Solution:**

We are given,

m = a cos

^{3}Î¸ + 3a cos Î¸ sin^{2}Î¸ and n = a sin^{3}Î¸ + 3a cos^{2}Î¸ sin Î¸So, L.H.S. =

= (a cos

^{3 }Î¸ + 3a cos Î¸ sin^{2}Î¸ + a sin^{3}Î¸ + 3a cos^{2}Î¸ sin Î¸)^{2/3}+ (a cos^{3}Î¸ + 3a cos Î¸ sin^{2}Î¸ â€“ a sin^{3}Î¸ â€“ 3a cos^{2}Î¸ sin Î¸)^{2/3}= a

^{2/3}((cos Î¸ + sin Î¸)^{3})^{2/3 }+ a^{2/3}((cos Î¸ âˆ’ sin Î¸)^{3})^{2/3 }= a

^{2/3}[(cos Î¸ + sin Î¸)^{2}+ (cos Î¸ âˆ’ sin Î¸)^{2}]= a

^{2/3}[cos^{2}Î¸ + sin^{2}Î¸ + 2 sin Î¸ cos Î¸ + cos^{2 }Î¸ + sin^{2}Î¸ âˆ’ 2 sin Î¸ cos Î¸]= 2 a

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

Hence proved.

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

^{3 }Î¸, y = b sin

^{3}Î¸, prove that (x/a)

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

^{2/3}= 1.

**Solution:**

Given x = a cos

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

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

= (cos

^{3}Î¸)^{2/3}+ (sin^{3}Î¸)^{2/3}= cos

^{2 }Î¸ + sin^{2}Î¸= 1

= R.H.S.

Hence proved.

**Question 78. If a cos Î¸ + b sin Î¸ = m and a sin Î¸ â€“ b cos Î¸ = 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 Î¸ + b sin Î¸)

^{2}+ (a sin Î¸ â€“ b cos Î¸)^{2}= a

^{2}cos^{2}Î¸ + b^{2}sin^{2}Î¸ + 2ab sin Î¸ cos Î¸ + a^{2 }sin^{2 }Î¸ + b^{2}cos^{2 }Î¸ â€“ 2ab sin Î¸ cos Î¸= a

^{2}cos^{2 }Î¸ + a^{2}cos^{2 }Î¸ + b^{2}sin^{2}Î¸ + b^{2 }cos^{2}Î¸= a

^{2 }(sin^{2}Î¸ + cos^{2}Î¸) + b^{2 }(sin^{2}Î¸ + cos^{2}Î¸)= 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 âˆ’ 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 Î¸ + cos**^{2} Î¸ = 1, prove that sin^{12} Î¸ + 3 sin^{10} Î¸ + 3 sin^{8} Î¸ + sin^{6} Î¸ + 2 sin^{4} Î¸ + 2 sin^{2} Î¸ âˆ’ 2 = 1.

^{2}Î¸ = 1, prove that sin

^{12}Î¸ + 3 sin

^{10}Î¸ + 3 sin

^{8}Î¸ + sin

^{6}Î¸ + 2 sin

^{4}Î¸ + 2 sin

^{2}Î¸ âˆ’ 2 = 1.

**Solution:**

We are given,

=> cos Î¸ + cos

^{2}Î¸ = 1=> cos Î¸ = 1 âˆ’ cos

^{2}Î¸=> cos Î¸ = sin

^{2}Î¸ . . . . (1)Now, L.H.S. = sin

^{12}Î¸ + 3 sin^{10}Î¸ + 3 sin^{8}Î¸ + sin^{6}Î¸ + 2 sin^{4}Î¸ + 2 sin^{2}Î¸ âˆ’ 2= (sin

^{4}Î¸)^{3}+ 3 sin^{4}Î¸ sin^{2}Î¸ (sin^{4}Î¸ + sin^{2}Î¸) + (sin^{2}Î¸)^{3}+ 2(sin^{2 }Î¸)^{2}+ 2 sin^{2}Î¸ âˆ’ 2Using (1), we get,

= (sin

^{4}Î¸ + sin^{2}Î¸)^{3}+ 2cos^{2}Î¸ + 2 cos Î¸ âˆ’ 2= ((sin

^{2}Î¸)^{2}+ sin^{2}Î¸)^{3}+ 2 cos^{2}Î¸ + 2 cos Î¸ â€“ 2= (cos

^{2}Î¸ + sin^{2}Î¸)^{3}+ 2 cos^{2}Î¸ + 2 cos Î¸ âˆ’ 2= 1 + 2(cos

^{2}Î¸ + sin^{2}Î¸) âˆ’ 2= 1 + 2(1) âˆ’2

= 1

= R.H.S.

Hence proved.

**Question 81. Given that: (1 + cos Î±)(1 + cos Î²)(1 + cos Î³) = (1 â€“ cos Î±)(1 â€“ cos Î²)(1 â€“ cos Î³). Show that one of the values of each member of this equality is sin Î± sin Î² sin Î³. **

**Solution:**

We have,

= (1 + cos Î±)(1 + cos Î²)(1 + cos Î³)

= 2 cos

^{2}(Î±/2).2 cos^{2}(Î²/2).2 cos^{2}(Î³/2)=

=

=

Therefore, sin Î± sin Î² sin Î³ is the member of equality.

Hence proved.

**Question 82. If sin Î¸ + cos Î¸ = x, prove that sin**^{6} Î¸ + cos^{6} Î¸ **= ****.**

^{6}Î¸ + cos

^{6}Î¸

**Solution:**

We are given,

=> sin Î¸ + cos Î¸ = x

On squaring both sides, we get,

=> (sin Î¸ + cos Î¸)

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

^{2 }Î¸ + cos^{2 }Î¸ + 2 sin Î¸ cos Î¸ = x^{2}=> 2 sin Î¸ cos Î¸ = x

^{2}âˆ’ 1=> sin Î¸ cos Î¸ = (x

^{2}âˆ’ 1)/2 . . . . (1)We know,

=> sin

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

=> (sin

^{2}Î¸ + cos^{2}Î¸)^{3}= 1^{3}=> sin

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

^{6}Î¸ + cos^{6}Î¸ = 1 â€“ 3 sin^{2}Î¸ cos^{2}Î¸From (1), we get,

=> sin

^{6}Î¸ + cos^{6}Î¸ = 1 â€“=> sin

^{6}Î¸ + cos^{6}Î¸ =

Hence proved.

**Question 83. If x = a sec Î¸ cos Ï•, y = b sec Î¸ sin Ï• and z = c tan Ï•, show that, x**^{2}/a^{2} + y^{2}/b^{2} âˆ’ z^{2}/c^{2} = 1.

^{2}/a

^{2}+ y

^{2}/b

^{2}âˆ’ z

^{2}/c

^{2}= 1.

**Solution:**

We are given, x = a sec Î¸ cos Ï•, y = b sec Î¸ sin Ï•, z = c tan Ï•

On squaring x, y, z, we get,

x

^{2 }= a^{2 }sec^{2 }Î¸ cos^{2}Ï• or x^{2}/a^{2}= sec^{2 }Î¸ cos^{2}Ï• . . . . (1)y

^{2}= b^{2}sec^{2 }Î¸ sin^{2}Ï• or y^{2}/b^{2}= sec^{2 }Î¸ sin^{2}Ï• . . . . (2)z

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

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

= sec

^{2}Î¸ cos^{2}Ï• + sec^{2}Î¸ sin^{2}Ï• âˆ’ tan^{2}Ï•= sec

^{2}Î¸ (cos^{2}Ï• + sin^{2}Ï•) âˆ’ tan^{2}Ï•= sec

^{2}Î¸ (1) âˆ’ tan^{2}Ï•= sec

^{2 }Î¸ âˆ’ tan^{2 }Î¸= 1

= R.H.S.

Hence proved.

**Question 84. If sin Î¸ + 2 cos Î¸. Prove that 2 sin Î¸ â€“ cos Î¸ = 2. **

**Solution:**

We are given, sin Î¸ + 2 cos Î¸ = 1

On squaring both sides, we get,

=> (sin Î¸ + 2 cos Î¸)

^{2}= 12=> sin

^{2 }Î¸ + 4 cos^{2 }Î¸ + 4 sin Î¸ cos Î¸ = 1=> 4 cos

^{2}Î¸ + 4 sin Î¸ cos Î¸ = 1 â€“ sin^{2 }Î¸=> 4 cos

^{2 }Î¸ + 4 sin Î¸ cos Î¸ â€“ cos^{2 }Î¸ = 0=> 3 cos

^{2 }Î¸ + 4 sin Î¸ cos Î¸ = 0 . . . . (1)We have, L.H.S. = 2 sin Î¸ â€“ cos Î¸

On squaring L.H.S., we get,

= (2 sin Î¸ â€“ cos Î¸)

^{2}= 4 sin

^{2 }Î¸ + cos^{2 }Î¸ â€“ 4 sin Î¸ cos Î¸From (1), we get,

= 4 sin

^{2 }Î¸ + cos^{2 }Î¸ + 3 cos^{2}Î¸= 4 sin

^{2}Î¸ + 4 cos^{2}Î¸= 4(sin

^{2 }Î¸ + cos^{2 }Î¸)= 4

So, we have,

=> (2 sin Î¸ â€“ cos Î¸)

^{2}= 4=> 2 sin Î¸ â€“ cos Î¸ = 2

Hence proved.

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