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• RD Sharma Class 10 Solutions

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

Question 57. tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B

Solution:

We have,

L.H.S. = tan2 A sec2 B − sec2 A tan2 B

= tan2 A (1 + tan2 B) − tan2 B (1+ tan2 A)

= tan2 A + tan2 A tan2 B − tan2 B − tan2 A tan2 B

= 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

= a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ − a2 tan2 θ − b2 sec2 θ – 2ab sec θ tan θ

= a2 sec2 θ + b2 tan2 θ − a2 tan2 θ − b2 sec2 θ

= a2 sec2 θ − b2 sec2 θ + b2 tan2θ − a2 tan2 θ

= sec2 θ (a2 − b2) + tan2 θ (b2 − a2)

= sec2θ (a2 − b2) − tan2θ (a2 − b2)

= (sec2 θ − 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)

= cosec2 θ − cot2 θ

= 1

= R.H.S.

Hence proved.

Question 61. If Tn = 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\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.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

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))

= (sec2 A − tan2 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.

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. sec4 A (1 − sin4 A) – 2 tan2 A = 1

Solution:

We have,

L.H.S. = sec4 A (1 − sin4 A) – 2 tan2 A

= sec4 A – tan4 A – 2 tan4 A

= (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 sin2 θ = m and a sin3 θ + 3a cos2 θ 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 cos3 θ + 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

= a2/3 ((cos θ + sin θ)3)2/3 + a2/3 ((cos θ − sin θ)3)2/3

= a2/3 [(cos θ + sin θ)2 + (cos θ − sin θ)2]

= a2/3 [cos2 θ + sin2 θ + 2 sin θ cos θ + cos2 θ + sin2 θ − 2 sin θ cos θ]

= 2 a2/3

= R.H.S.

Hence proved.

Question 77. If x = a cos3 θ, y = b sin3θ, prove that (x/a)2/3 + (y/b)2/3 = 1.

Solution:

Given x = a cos3 θ and y = b sin3 θ.

So, L.H.S. = (x/a)2/3 + (y/b)2/3

= (cos3 θ)2/3 + (sin3 θ)2/3

= cos2 θ + 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 + a2 sin2 θ + b2 cos2 θ – 2ab sin θ cos θ

= a2 cos2 θ + a2 cos2 θ + b2 sin2 θ + b2 cos2 θ

= a2 (sin2 θ + cos2 θ) + b2 (sin2 θ + cos2 θ)

= a2 + b

= L.H.S.

Hence proved.

Question 79. If cos A + cos2 A = 1, Prove that sin2 A + sin4 A = 1.

Solution:

We are given,

=> cos A + cos2 A = 1

=> cos A = 1 − cos2 A

=> 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(sin2 θ)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

=> sin2 θ + cos2 θ + 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 = a2 sec2 θ cos2 ϕ or x2/a2 = sec2 θ cos2 ϕ  . . . . (1)

y2 = b2 sec2 θ sin2 ϕ or y2/b2 = sec2 θ 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

= sec2 θ − tan2 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

=> sin2 θ + 4 cos2 θ + 4 sin θ cos θ = 1

=> 4 cos2 θ + 4 sin θ cos θ = 1 – sin2 θ

=> 4 cos2 θ + 4 sin θ cos θ – cos2 θ = 0

=> 3 cos2 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 sin2 θ + cos2 θ – 4 sin θ cos θ

From (1), we get,

= 4 sin2 θ + cos2 θ + 3 cos2θ

= 4 sin2 θ + 4 cos2 θ

= 4(sin2 θ + cos2 θ)

= 4

So, we have,

=> (2 sin θ – cos θ)2 = 4

=> 2 sin θ – cos θ  = 2

Hence proved.

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