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

  • Last Updated : 30 Apr, 2021

Prove the following trigonometric identities:

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 = \frac{5(1−cosθ)(1+cosθ)}{1+cosθ}

=> 3 sin θ = \frac{5(1−cos^2θ)}{1+cosθ}

=> 3 sin θ = \frac{5sin^2θ}{1+cosθ}

=> 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 \frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}.

Solution:

We have,

L.H.S. = \frac{T_3-T_5}{T_1}

\frac{(sin^3θ+cos^3θ)-(sin^5θ+cos^5θ)}{sinθ+cosθ}

\frac{sin^3θ(1-sin^2θ)+cos^3θ(1-cos^2θ)}{sinθ+cosθ}

\frac{sin^3θ(cos^2θ)+cos^3θ(sin^2θ)}{sinθ+cosθ}

\frac{sin^2θcos^2θ(sinθ+cosθ)}{sinθ+cosθ}

= sin2 θ cos2 θ

And R.H.S. = \frac{T_5-T_7}{T_3}

\frac{(sin^5θ+cos^5θ)-(sin^7θ+cos^7θ)}{sin^3θ+cos^3θ}

\frac{sin^5θ(1-sin^2θ)+cos^5θ(1-cos^2θ)}{sin^3θ+cos^3θ}

\frac{sin^5θ(cos^2θ)+cos^5θ(sin^2θ)}{sin^3θ+cos^3θ}

\frac{sin^2θcos^2θ(sin^3θ+cos^3θ)}{sin^3θ+cos^3θ}

= sin2 θ cos2 θ

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

Hence proved.

Question 62. (tanθ+\frac{1}{cosθ})^2+(tanθ-\frac{1}{cosθ})^2=2(\frac{1+sin^2θ}{1-sin^2θ})

Solution:

We have,

L.H.S. = (tanθ+\frac{1}{cosθ})^2+(tanθ-\frac{1}{cosθ})^2

= (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.

Question 63. (\frac{1}{sec^2θ-cos^2θ}+\frac{1}{cosec^2θ-sin^2θ})sin^2θcos^2θ=\frac{1-sin^2θcos^2θ}{2+sin^2θcos^2θ}

Solution:

We have,

L.H.S. = (\frac{1}{sec^2θ-cos^2θ}+\frac{1}{cosec^2θ-sin^2θ})sin^2θcos^2θ

(\frac{1}{\frac{1}{cos^2θ}-cos^2θ}+\frac{1}{\frac{1}{sin^2θ}-sin^2θ})sin^2θcos^2θ

(\frac{cos^2θ}{1-cos^4θ}+\frac{sin^2θ}{1-sin^4θ})sin^2θcos^2θ

(\frac{cos^2θ}{cos^2θ+sin^2θ-cos^4θ}+\frac{sin^2θ}{sin^2θ+cos^2θ-sin^4θ})sin^2θcos^2θ

[\frac{cos^2θ}{cos^2θ(1-cos^2θ)+sin^2θ}+\frac{sin^2θ}{sin^2θ(1-sin^2θ)+cos^2θ}]sin^2θcos^2θ

[\frac{cos^2θ}{sin^2θ(cos^2θ+1)}+\frac{sin^2θ}{cos^2θ(sin^2θ+1)}]sin^2θcos^2θ

\left[\frac{cos^4θ+cos^4θsin^2θ+sin^4θ+sin^4θcos^2θ}{sin^2θcos^2θ(cos^2θ+1)(sin^2θ+1)}\right]sin^2θcos^2θ

\frac{1-2sin^2θcos^2θ+sin^2θcos^2θ(cos^2θ+sin^2θ)}{1+cos^2θ+sin^2θ+cos^2θsin^2θ}

\frac{1-sin^2θcos^2θ}{2+cos^2θsin^2θ}

= R.H.S.

Hence proved.

Question 64. (i) \left[\frac{1+sinθ-cosθ}{1+sinθ+cosθ}\right]^2=\frac{1-cosθ}{1+cosθ}

Solution:

We have,

L.H.S. = \left[\frac{1+sinθ-cosθ}{1+sinθ+cosθ}\right]^2

\left[\frac{(1+sinθ-cosθ)(1+sinθ-cosθ)}{(1+sinθ+cosθ)(1+sinθ-cosθ)}\right]^2

\left[\frac{(1+sinθ-cosθ)^2}{(1+sinθ)^2-cos^2θ}\right]^2

\left[\frac{1+sin^2θ+cos^2θ+2sinθ-2sinθcosθ-2cosθ}{1+sin^2θ+2sinθ-cos^2θ}\right]^2

\left[\frac{2+2sinθ-2sinθcosθ-2cosθ}{2sin^2θ+2sinθ}\right]^2

\left[\frac{2(1+sinθ)-2cosθ(sinθ+1)}{2sinθ(sinθ+1)}\right]^2

\left[\frac{(1+sinθ)(2-2cosθ)}{2sinθ(sinθ+1)}\right]^2

\left[\frac{2-2cosθ}{2sinθ}\right]^2

\frac{(1-cosθ)(1-cosθ)}{sin^2θ}

\frac{(1-cosθ)(1-cosθ)}{1-cos^2θ}

\frac{(1-cosθ)(1-cosθ)}{(1+cosθ)(1-cosθ)}

\frac{1-cosθ}{1+cosθ}

= R.H.S.

Hence proved.

(ii) \frac{1+secθ-tanθ}{1+secθ+tanθ}=\frac{1-sinθ}{cosθ}

Solution:

We have,

L.H.S. = \frac{1+secθ-tanθ}{1+secθ+tanθ}

\frac{(sec^2θ-tan^2θ)+secθ-tanθ}{1+secθ+tanθ}

\frac{(secθ+tanθ)(secθ-tanθ)+(secθ-tanθ)}{1+secθ+tanθ}

\frac{(secθ-tanθ)[1+secθ+tanθ]}{1+secθ+tanθ}

= sec θ − tan θ

= 1/cos θ − sin θ/cos θ

\frac{1-sinθ}{cosθ}

= 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)

(\frac{cosA-1+sinA}{cosA})(\frac{cosA+1+sinA}{cosA})

\frac{(cosA+sinA)^2-1}{cos^2A}

\frac{cos^2A+sin^2A+2cosAsinA-1}{cos^2A}

\frac{2cosAsinA}{cos^2A}

= 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)

(\frac{sinA+cosA-1}{sinA})(\frac{cosA+sinA+1}{cosA})

\frac{(sinA+cosA)^2-1}{sinAcosA}

\frac{sin^2A+cos^2A+2sinAcosA-1}{sinAcosA}

\frac{2sinAcosA}{sinAcosA}

= 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)

[\frac{1}{sinθ}-\frac{1}{cosθ}][\frac{cosθ}{sinθ}-\frac{sinθ}{cosθ}]

[\frac{cosθ-sinθ}{sinθcosθ}][\frac{cos^2θ-sin^2θ}{sinθcosθ}]

\left[\frac{(cosθ-sinθ)^2(cosθ+sinθ)}{sin^2θcos^2θ}\right]

And R.H.S. = (cosec θ + sec θ) (sec θ cosec θ − 2) 

[\frac{1}{sinθ}+\frac{1}{cosθ}][(\frac{1}{cosθ})(\frac{1}{sinθ})-2]

[\frac{sinθ+cosθ}{sinθcosθ}][\frac{1-2sinθcosθ}{sinθcosθ}]

[\frac{sinθ+cosθ}{sinθcosθ}][\frac{sin^2θ+cos^2θ-2sinθcosθ}{sinθcosθ}]

[\frac{sinθ+cosθ}{sinθcosθ}][\frac{(sinθ-cosθ)^2}{sinθcosθ}]

\left[\frac{(cosθ-sinθ)^2(cosθ+sinθ)}{sin^2θcos^2θ}\right]

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

Hence proved.

Question 68. \frac{cosAcosecA-sinAsecA}{cosA+sinA}=cosecA-secA

Solution:

We have,

L.H.S. = \frac{cosAcosecA-sinAsecA}{cosA+sinA}

\frac{\frac{cosA}{sinA}-\frac{sinA}{cosA}}{cosA+sinA}

\frac{cos^2A-sin^2A}{cosAsinA}\left(\frac{1}{cosA+sinA}\right)

\frac{cosA-sinA}{cosAsinA}

= cosec A − sec A

= R.H.S.

Hence proved.

Question 69. \frac{sinA}{secA+tanA-1}+\frac{cosA}{cosecA+cotA-1}=1

Solution:

We have,

L.H.S. = \frac{sinA}{secA+tanA-1}+\frac{cosA}{cosecA+cotA-1}

\frac{sinA}{\frac{1+sinA-cosA}{cosA}}+\frac{cosA}{\frac{1+cosA-sinA}{sinA}}

sinAcosA(\frac{1}{1+sinA-cosA}+\frac{1}{1+cosA-sinA})

\frac{2sinAcosA}{(1+sinA-cosA)(1+cosA-sinA)}

\frac{2sinAcosA}{cosA-sinA+sinA+sinAcosA-sin^2A-cosA-cos^2A+cosAsinA}

\frac{2sinAcosA}{1-(sin^2A+cos^2A)+2sinAcosA}

\frac{2sinAcosA}{2sinAcosA}

= 1

= R.H.S.

Hence proved.

Question 70. \frac{tanA}{(1+tan^2A)^2}+\frac{cotA}{(1+cot^2A)^2}=sinAcosA

Solution:

We have,

\frac{tanA}{(1+tan^2A)^2}+\frac{cotA}{(1+cot^2A)^2}

\frac{tanA}{sec^4A}+\frac{cotA}{cosec^4A}

\frac{\frac{sinA}{cosA}}{\frac{1}{cos^4A}}+\frac{\frac{cosA}{sinA}}{\frac{1}{sin^4A}}

= 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. \frac{cot^2A(secA-1)}{1+sinA}=sec^2A(\frac{1-sinA}{1+secA})

Solution:

We have,

L.H.S. = \frac{cot^2A(secA-1)}{1+sinA}

\frac{\frac{cos^2A}{sin^2A}(\frac{1-cosA}{cosA})}{1+sinA}

\frac{\frac{cosA(1-cosA)}{1-cos^2A}}{1+sinA}

\frac{cosA}{(1+sinA)(1+cosA)}

And R.H.S. = sec^2A(\frac{1-sinA}{1+secA})   

\frac{1}{cos^2A}(\frac{1-sinA}{1+secA})

\frac{cosA}{cos^2A}(\frac{1-sinA}{1+cosA})

\frac{1}{cosA}(\frac{1-sinA}{1+cosA})

\frac{1+sinA}{cosA(1+sinA)}(\frac{1-sinA}{1+cosA})

\frac{1-sin^2A}{cosA(1+sinA)(1+cosA)}

\frac{cos^2A}{cosA(1+sinA)(1+cosA)}

\frac{cosA}{(1+sinA)(1+cosA)}

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

=> \frac{x^2cos^2θ}{a^2}+\frac{y^2sin^2θ}{b^2}+\frac{2xycosθsinθ}{ab}+\frac{x^2sin^2θ}{a^2}+\frac{y^2cos^2θ}{b^2}-\frac{2xycosθsinθ}{ab}   = 2

=> cos^2θ(\frac{x^2}{a^2}+\frac{y^2}{b^2})+sin^2θ(\frac{x^2}{a^2}+\frac{y^2}{b^2})   = 2

=> \frac{x^2}{a^2}+\frac{y^2}{b^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\frac{1-sin^2θ}{sinθ}

=> a3\frac{cos^2θ}{sinθ}

=> a = \frac{cos^\frac{2}{3}θ}{sin^\frac{1}{3}θ}

On squaring both sides, we get,

=> a2\frac{cos^\frac{4}{3}θ}{sin^\frac{2}{3}θ}

Also we have,

=> sec θ – cos θ = b3

=> 1/cos θ – cos θ = b3

=> b3\frac{1-cos^2θ}{cosθ}

=> b3\frac{sin^2θ}{cosθ}

=> b = \frac{sin^\frac{2}{3}θ}{cos^\frac{1}{3}θ}

On squaring both sides, we get,

=> b2\frac{sin^\frac{4}{3}θ}{cos^\frac{2}{3}θ}

So, L.H.S. = a2b2 (a2+ b2)

\frac{cos^\frac{4}{3}θ}{sin^\frac{2}{3}θ}×\frac{sin^\frac{4}{3}θ}{cos^\frac{2}{3}θ}\left(\frac{cos^\frac{4}{3}θ}{sin^\frac{2}{3}θ}+\frac{sin^\frac{4}{3}θ}{cos^\frac{2}{3}θ}\right)

cos^{\frac{2}{3}}θsin^{\frac{2}{3}}θ\left(\frac{1}{cos^{\frac{2}{3}}θsin^{\frac{2}{3}}θ}\right)

= 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  

(m + n)^{\frac{2}{3}} + (m − n)^{\frac{2}{3}} = 2a^{\frac{2}{3}}

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. = (m + n)^{\frac{2}{3}} + (m − n)^{\frac{2}{3}}

= (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

\left(\frac{acos^3θ}{a}\right)^{\frac{2}{3}}+\left(\frac{bcos^3θ}{b}\right)^{\frac{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)

\left(\frac{8cos^2(\frac{α}{2})cos^2(\frac{β}{2})cos^2(\frac{γ}{2})}{sinα.sinβ.sinγ}\right)sinα.sinβ.sinγ

\left(\frac{8sin^2(\frac{α}{2})sin^2(\frac{β}{2})sin^2(\frac{γ}{2})}{2sin(\frac{α}{2})cos(\frac{α}{2}).sin(\frac{β}{2})cos(\frac{β}{2}).sin(\frac{γ}{2})cos(\frac{γ}{2})}\right)sinα.sinβ.sinγ

sinα.sinβ.sinγ×tan\frac{α}{2}.tan\frac{β}{2}.tan\frac{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 \frac{4-3(x^2-1)^2}{4}.

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 – \frac{3(x^2-1)^2}{4}

=> sin6 θ + cos6 θ = \frac{4-3(x^2-1)^2}{4}

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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