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Class 10 RD Sharma Solutions – Chapter 6 Trigonometric Identities – Exercise 6.2

  • Last Updated : 28 Apr, 2021

Question 1. If cos θ = 4/5, find all other trigonometric ratios of angle θ.  

Solution:

We are given, cos θ = 4/5. So, sec θ =1/cos θ = 5/4.

Now we know,

=> sin θ = √(1 – cos2 θ)

=> sin θ = √(1 – (4/5)2)



=> sin θ = √(1 – (16/25))

=> sin θ = √(9/25)

=> sin θ = 3/5

So, cosec θ = 1/sin θ = 5/3

And tan θ = sin θ/cos θ = (3/5)/(4/5) = 3/4

Therefore, cot θ = 1/tan θ = 4/3

If cos θ = 4/5, value of sec θ, sin θ, cosec θ, tan θ and cot θ are 5/4, 3/5, 5/3, 3/4 and 4/3 respectively.

Question 2. If sin θ = 1/√2, find all other trigonometric ratios of angle θ.

Solution:



We are given, sin θ = 1/√2. So, cosec θ =1/sin θ = √2.

Now we know,

=> cos θ = √(1 – sin2 θ)

=> cos θ = √(1 – (1/√2)2)

=> cos θ = √(1 – (1/2))

=> cos θ = √(1/2)

=> cos θ = 1/√2

So, sec θ = 1/cos θ = √2

And tan θ = sin θ/cos θ = (1/√2)/(1/√2) = 1

Therefore, cot θ = 1/tan θ = 1

If sin θ = 1/√2, value of cosec θ, cos θ, sec θ, tan θ and cot θ are √2, 1/√2, √2, 1 and 1 respectively.

Question 3. If tan θ = 1/√2, find the value of \frac{cosec^2θ-sec^2θ}{cosec^2θ+cot^2θ}.

Solution:

We are given, tan θ = 1/√2. Now we know,

=> sec θ = √(1 + tan2 θ)

=> sec θ = √(1 + (1/√2)2)

=> sec θ = √(1+(1/2))

=> sec θ = √(3/2)

And cot θ = 1/tan θ = √2. Also, we know,

=> cosec θ = √(1 + cot2 θ)

=> cosec θ = √(1 + (√2)2)



=> cosec θ = √(1 + 2)

=> cosec θ = √3

So,  \frac{cosec^2θ-sec^2θ}{cosec^2θ+cot^2θ}   = \frac{(\sqrt{3})^2-\left(\sqrt{\frac{3}{2}}\right)^2}{(\sqrt{3})^2+\left(\sqrt{2}\right)^2}

\frac{3-\frac{3}{2}}{3+2}

\frac{3}{10}

Therefore, the value of \frac{cosec^2θ-sec^2θ}{cosec^2θ+cot^2θ}  is \frac{3}{10}.

Question 4. If tan θ = 3/4, find the value of \frac{1-cosθ}{1+cosθ}.

Solution:

We are given, tan θ = 3/4. Now we know,

=> sec θ = √(1 + tan2 θ)

=> sec θ = √(1 + (3/4)2)



=> sec θ = √(1+(9/16))

=> sec θ = √(25/16)

=> sec θ = 5/4

And cos θ = 1/sec θ = 4/5.

So, \frac{1-cosθ}{1+cosθ}   = \frac{1-\frac{4}{5}}{1+\frac{4}{5}}

\frac{\frac{1}{5}}{\frac{9}{5}}

\frac{1}{9}

Therefore, the value of \frac{1-cosθ}{1+cosθ}  is \frac{1}{9}.

Question 5. If tan θ = 12/5, find the value of \frac{1+sinθ}{1-sinθ}.

Solution:

We are given, tan θ = 12/5. So, cot θ = 1/tan θ = 5/12.



Now we know,

=> cosec θ = √(1 + cot2 θ)

=> cosec θ = √(1 + (5/12)2)

=> cosec θ = √(1 + (25/144))

=> cosec θ = √(169/144)

=> cosec θ = 13/12

And sin θ = 1/cosec θ = 12/13.

So, \frac{1+sinθ}{1-sinθ}   = \frac{1+\frac{12}{13}}{1-\frac{12}{13}}

\frac{\frac{25}{13}}{\frac{1}{13}}

= 25

Therefore, the value of \frac{1+sinθ}{1-sinθ}  is 25.

Question 6. If cot θ = 1/√3, find the value of \frac{1-cos^2θ}{2-sin^2θ}.

Solution:

We are given, cot θ = 1/√3. Now we know,

=> cosec θ = √(1 + cot2 θ)

=> cosec θ = √(1 + (1/√3)2)

=> cosec θ = √(1 + (1/3))

=> cosec θ = √(4/3)

=> cosec θ = 2/√3

And sin θ = 1/cosec θ = √3/2. Also, we know,

=> cos θ = √(1 – sin2 θ)



=> cos θ = √(1 – (√3/2)2)

=> cos θ = √(1 – (3/4))

=> cos θ = √(1/4)

=> cos θ = 1/2

So, \frac{1-cos^2θ}{2-sin^2θ}   = \frac{1-(\frac{1}{2})^2}{2-(\frac{\sqrt{3}}{2})^2}

\frac{1-\frac{1}{4}}{2-\frac{3}{4}}

\frac{\frac{3}{4}}{\frac{5}{4}}

\frac{3}{5}

Therefore, the value of \frac{1-cos^2θ}{2-sin^2θ}  is \frac{3}{5}.

Question 7. If cosec A = √2, find the value of \frac{2sin^2A+3cot^2A}{4(tan^2A-cos^2A)}.

Solution:



We are given, cosec A = √2. So, sin A = 1/cosec A = 1/√2.

Now we know,

=> cos A = √(1 – sin2 A)

=> cos A = √(1 – (1/√2)2)

=> cos A = √(1 – (1/2))

=> cos A = √(1/2)

=> cos A = 1/√2

Hence, tan A = sin A/cos A = (1/√2)/(1/√2) = 1. And cot A = 1/tan A = 1.

So, \frac{2sin^2A+3cot^2A}{4(tan^2A-cos^2A)}   = \frac{2(\frac{1}{\sqrt{2}})^2+3(1)^2}{4\left(1^2-(\frac{1}{\sqrt{2}})^2\right)}

\frac{1+3}{2}

= 2

Therefore, the value of \frac{2sin^2A+3cot^2A}{4(tan^2A-cos^2A)}  is 2.

Question 8. If cot θ = √3, find the value of \frac{cosec^2θ+cot^2θ}{cosec^2θ-sec^2θ}.

Solution:

We are given cot θ = √3. And tan θ = 1/cot θ = 1/√3.

Now we know,

=> cosec θ = √(1 + cot2 θ)

=> cosec θ = √(1 + (√3)2)

=> cosec θ = √4

=> cosec θ = 2

Also, we know,



=> sec θ = √(1 + tan2 θ)

=> sec θ = √(1 + (1/√3)2)

=> sec θ = √(1+(1/3))

=> sec θ = √(4/3)

=> sec θ = 2/√3

So, \frac{cosec^2θ+cot^2θ}{cosec^2θ-sec^2θ}   = \frac{2^2+(\sqrt{3})^2}{2^2-\left(\frac{2}{\sqrt{3}}\right)^2}

\frac{7}{\frac{8}{3}}

\frac{21}{8}

Therefore, the value of \frac{cosec^2θ+cot^2θ}{cosec^2θ-sec^2θ} is \frac{21}{8}.

Question 9. If 3cos θ = 1, find the value of \frac{6sin^2θ+tan^2θ}{4cosθ}.

Solution:

We are given cos θ = 1/3. Now we know,

=> sin θ = √(1 – cos2 θ)

=> sin θ = √(1 – (1/3)2)

=> sin θ = √(1 – (1/9))

=> sin θ = √(8/9)

=> sin θ = 2√2/3

Hence, tan θ = sin θ/cos θ = (2√2/3)/(1/3) = 2√2

So, \frac{6sin^2θ+tan^2θ}{4cosθ}   = \frac{6(\frac{2\sqrt{2}}{3})^2+(2\sqrt{2})^2}{4(\frac{1}{3})}

\frac{\frac{16}{3}+8}{\frac{4}{3}}

\frac{40}{4}



= 10

Therefore, the value of \frac{6sin^2θ+tan^2θ}{4cosθ}  is 10.

Question 10. If √3 tan θ = sin θ, find the value of sin2 θ – cos2 θ. 

Solution:

We are given, √3 tan θ = sin θ

=> √3 (sin θ/cos θ) = sin θ

=> cos θ = 1/√3

Now we know,

=> sin θ = √(1 – cos2 θ)

=> sin θ = √(1 – (1/√3)2)

=> sin θ = √(1 – (1/3))

=> sin θ = √(2/3)

So, sin2 θ – cos2 θ = √(2/3)2 – (1/√3)2

= 2/3 – 1/3 

= 1/3 

Therefore, the value of sin2 θ – cos2 θ is 1/3.

Question 11. If cosec θ = 13/12, find the value of \frac{2sinθ-3cosθ}{4sinθ-9cosθ}.

Solution:

We are given, cosec θ = 13/12. So, sin θ = 1/cosec θ = 12/13.

Now we know,

=> cos θ = √(1 – sin2 θ)

=> cos θ = √(1 – (12/13)2)

=> cos θ = √(1 – (144/169))

=> cos θ = √(25/169)

=> cos θ = 5/13

So, \frac{2sinθ-3cosθ}{4sinθ-9cosθ}   = \frac{2(\frac{12}{13})-3(\frac{5}{13})}{4(\frac{12}{13})-9(\frac{5}{13})}

\frac{\frac{24-15}{13}}{\frac{48-45}{13}}

\frac{9}{3}

= 3

Therefore, the value of \frac{2sinθ-3cosθ}{4sinθ-9cosθ} is 3.

Question 12. If sin θ + cos θ = √2 cos (90o–θ), find cot θ.

Solution:

We are given,



=> sin θ + cos θ = √2 cos (90o–θ)

=> sin θ + cos θ = √2 sin θ 

=> cos θ = (√2–1)sin θ

=> cos θ/sin θ = √2–1 

=> cot θ = √2–1 

Therefore, value of cot θ is √2–1.  

Question 13. If 2sin2 θ – cos2 θ = 2, then find the value of θ.

Solution:

We have,

=> 2sin2 θ – cos2 θ = 2

=> 2sin2 θ – (1 – sin2 θ) = 2

=> 2sin2 θ – 1 + sin2 θ = 2

=> 3sin2 θ = 3

=> sin2 θ = 1

=> sin θ = 1

=> θ = 90o

Therefore, the value of θ is 90o.

Question 14. If √3tan θ – 1 = 0, find the value of sin2 θ – cos2 θ. 

Solution:

We are given,

=> √3tan θ – 1 = 0

=> tan θ = 1/√3

=> tan θ = tan 30o

=> θ = 30o

So, sin2 θ – cos2 θ = sin2 30o – cos2 30o

= (1/2)2 – (√3/2)2

= (1/4) – (3/4)

= –2/4

= –1/2

Therefore, the value of sin2 θ – cos2 θ is –1/2.

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