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Class 10 RD Sharma Solutions – Chapter 6 Trigonometric Identities – Exercise 6.1 | Set 1

  • Last Updated : 30 Apr, 2021

Prove the following trigonometric identities:

Question 1. (1 – cos2 A) cosec2 A = 1

Solution:

We have, 

L.H.S. = (1 – cos2 A) cosec2 A

By using the identity, sin2 A + cos2 A = 1, we get,

= (sin2 A) (cosec2 A)



= sin2 A × (1/sin2 A)

= 1 

= R.H.S.

Hence proved.

Question 2. (1 + cot2 A) sin2 A = 1  

Solution: 

We have,

L.H.S. = (1 + cot2 A) sin2 A

By using the identity, cosec2 A = 1 + cot2 A, we get,

= cosec2 A sin2 A

= (1/sin2 A) × sin2 A

= 1

= R.H.S

Hence proved.

Question 3. tan2 θ cos2 θ = 1 − cos2 θ  

Solution: 

We have,

L.H.S. = tan2 θ cos2 θ

= (sin2 θ/cos2 θ) (cos2 θ)

= sin2 θ



= 1 − cos2 θ  

= R.H.S.

Hence proved.

Question 4. cosec θ √(1 – cos2 θ) = 1

Solution:

We have,

L.H.S. = cosec θ √(1 – cos2 θ)

= cosec θ √(sin2 θ)

= cosec θ sin θ

= (1/sin θ) (sin θ)

= 1

= R.H.S.

Hence proved.

Question 5. (sec2 θ − 1)(cosec2 θ − 1) = 1  

Solution:

We have,

L.H.S. = (sec2 θ − 1)(cosec2 θ − 1)

By using the identities sec2 θ − tan2 θ = 1 and cosec2 θ − cot2 θ = 1, we have,

= tan2 θ cot2 θ

= (tan2 θ) (1/tan2 θ)

= 1

= R.H.S.



Hence proved.

Question 6. tan θ + 1/tan θ = sec θ cosec θ

Solution:

We have,

L.H.S. = tan θ + 1/ tan θ

= (tan2 θ + 1)/tan θ

= sec2 θ/tan θ

\frac{\frac{1}{cos^2θ}}{\frac{1}{\frac{sinθ}{cosθ}}}

\frac{cos θ}{sinθcos^2 θ}

= 1/sin θ cos θ

= sec θ cosec θ



= R.H.S.

Hence proved.

Question 7. cos θ/(1 – sin θ) = (1 + sin θ)/cos θ

Solution:

We have,

L.H.S. = cos θ/(1 – sin θ)

\frac{cosθ(1+sin θ)}{(1-sinθ)(1+sin θ)}      

\frac{cosθ(1+sin θ)}{(1-sin^2θ)}

\frac{cosθ(1+sinθ)}{cos^2θ}

= (1 + sin θ)/cos θ

= R.H.S.

Hence proved.

Question 8. cos θ/(1 + sin θ) = (1 –  sin θ)/cos θ

Solution:

We have,

L.H.S. = cos θ/(1 + sin θ)

\frac{cosθ(1-sin θ)}{(1+sinθ)(1-sin θ)}

\frac{cosθ(1-sin θ)}{(1-sin^2θ)}

\frac{cosθ(1-sinθ)}{cos^2θ}

= (1 –  sin θ)/cos θ

= R.H.S.

Hence proved.



Question 9. cos2 θ + 1/(1 + cot2 θ) = 1

Solution:

We have, 

L.H.S. = cos2 θ + 1/(1 + cot2 θ)

= cos2 θ + 1/(cosec2 θ)

= cos2 θ + sin2 θ

= 1

= R.H.S.

Hence proved.

Question 10. sin2 A + 1/(1 + tan2 A) = 1

Solution:

We have,

L.H.S. = sin2 A + 1/(1 + tan2 A)

= sin2 A + 1/(sec2 A)

= sin2 A + cos2 A

= 1

= R.H.S.

Hence proved.

Question 11. \sqrt{\frac{1-cosθ}{1+cosθ}} = cosec θ − cot θ

Solution:

We have,

L.H.S. = \sqrt{\frac{1-cosθ}{1+cosθ}}     

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



\sqrt{\frac{(1-cosθ)^2}{1-cos^2θ}}

\sqrt{\frac{(1-cosθ)^2}{sin^2θ}}

\frac{1-cosθ}{sinθ}

\frac{1}{sinθ}-\frac{cosθ}{sinθ}

= cosec θ − cot θ

= R.H.S.

Hence proved.

Question 12. (1 – cos θ)/sin θ = sin θ/(1 + cos θ)

Solution:

We have,

L.H.S. = (1 – cos θ)/sin θ

\frac{(1-cosθ)(1+cosθ)}{sinθ(1+cosθ)}

\frac{1-cos^2θ}{sinθ(1+cosθ)}     

\frac{sin^2θ}{sinθ(1+cosθ)}

= sin θ/(1 + cos θ)

= R.H.S.

Hence proved.

Question 13. sin θ/(1 – cos θ) = cosec θ + cot θ

Solution:

We have,

L.H.S. = sin θ/(1 – cos θ)

\frac{sinθ(1+cosθ)}{(1-cosθ)(1+cosθ)}



\frac{sinθ(1+cosθ)}{(1-cos^2θ)}

\frac{sinθ(1+cosθ)}{sin^2θ}

\frac{1+cosθ}{sinθ}

\frac{1}{sinθ}+\frac{cosθ}{sinθ}

= cosec θ + cot θ

= R.H.S.

Hence proved.

Question 14. (1 – sin θ)/(1 + sin θ) = (sec θ – tan θ)2

Solution:

We have,

L.H.S. = (1 – sin θ)/(1 + sin θ)

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

\frac{(1-sinθ)^2}{(1-sin^2θ)}

\frac{(1-sinθ)^2}{cos^2θ}

\left(\frac{1-sinθ}{cosθ}\right)^2

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

= (sec θ – tan θ)2

= R.H.S.

Hence proved.

Question 15. \frac{(1+cot^2θ)tanθ}{sec^2θ}=cotθ

Solution:

We have,



L.H.S. = \frac{(1+cot^2θ)tanθ}{sec^2θ}

\frac{cosec^2θ×tanθ}{sec^2θ}

\frac{1}{sin^2θ}×\frac{cos^2θ}{1}×\frac{sinθ}{cosθ}

= cos θ/sin θ

= cot θ

= R.H.S.

Hence proved.

Question 16. tan2 θ − sin2 θ = tan2 θ sin2 θ 

Solution:

We have,

L.H.S. = tan2 θ − sin2 θ 

= sin2 θ/cos2 θ − sin2 θ

= sin2 θ(1/cos2 θ − 1)

sin^2θ(\frac{1-cos^2θ}{cos^2θ})

= sin2θ (sin2θ/cos2θ)

= tan2 θ sin2 θ 

= R.H.S.

Hence proved.

Question 17. (cosec θ + sin θ)(cosec θ – sin θ) = cot2θ + cos2θ 

Solution:

We have,

L.H.S. = (cosec θ + sin θ)(cosec θ – sin θ)

= cosec2 θ – sin2 θ

= (1 + cot2 θ) – (1 – cos2 θ)

= 1 + cot2 θ – 1 + cos2 θ

= cot2 θ + cos2 θ

= R.H.S.

Hence proved.

Question 18. (sec θ + cos θ) (sec θ – cos θ) = tan2 θ + sin2 θ 

Solution:

We have,

L.H.S. = (sec θ + cos θ) (sec θ – cos θ)

= sec2 θ – cos2 θ

= (1 + tan2 θ) – (1 – sin2 θ)

= 1 + tan2 θ – 1 + sin2 θ

= tan2 θ + sin2 θ

= R.H.S

Hence proved.

Question 19. sec A(1 – sin A) (sec A + tan A) = 1

Solution:

We have,

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

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

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



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

= 1

= R.H.S

Hence proved.

Question 20. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1 

Solution:

We have,

L.H.S. = (cosec A – sin A)(sec A – cos A)(tan A + cot A)

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

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

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

= 1

= R.H.S

Hence proved.

Question 21. (1 + tan2 θ)(1 – sin θ)(1 + sin θ) = 1

Solution:

We have,

L.H.S. = (1 + tan2 θ)(1 – sin θ)(1 + sin θ)

= (sec2 θ) (1 – sin2 θ)

= (sec2 θ) (cos2 θ)

= 1

= R.H.S

Hence proved.

Question 22. sin2 A cot2 A + cos2 A tan2 A = 1

Solution:

We have,

L.H.S. = sin2 A cot2 A + cos2 A tan2 A

= sin2 A (cos2 A/sin2 A) + cos2 A (sin2 A/cos2 A)

= cos2 A + sin2 A

= 1

= R.H.S.

Hence proved.

Question 23. 

(i) cot θ – tan θ = \frac{2cos^2θ-1}{sinθcosθ}

Solution:



We have,

L.H.S. = cot θ – tan θ

= cos θ/sin θ – sin θ/cos θ

\frac{cos^2θ-sin^2θ}{sinθcosθ}

\frac{cos^2θ-(1-cos^2θ)}{sinθcosθ}

\frac{cos^2θ-1+cos^2θ}{sinθcosθ}

\frac{2cos^2θ-1}{sinθcosθ}

= R.H.S.

Hence proved.

(ii) tan θ – cot θ = \frac{2sin^2θ-1}{sinθcosθ}

Solution:

We have,

L.H.S. = tan θ – cot θ

= sin θ/cos θ – cos θ/sin θ

\frac{sin^2θ-cos^2θ}{sinθcosθ}

\frac{sin^2θ-(1-sin^2θ)}{sinθcosθ}

\frac{sin^2θ-1+sin^2θ}{sinθcosθ}

\frac{2sin^2θ-1}{sinθcosθ}

= R.H.S.

Hence proved.

Question 24. (cos2 θ/sin θ) – cosec θ + sin θ = 0

Solution:



We have,

L.H.S. = (cos2 θ/sin θ) – cosec θ + sin θ

\frac{cos^2θ}{sinθ}-\frac{1}{sinθ}+sin θ

\frac{(cos^2θ-1)+sin^2θ}{sinθ}

\frac{-sin^2θ+sin^2θ}{sinθ}

= 0

= R.H.S.

Hence proved.

Question 25. \frac{1}{1+sinA}+\frac{1}{1-sinA}=2sec^2A

Solution:

We have,

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

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

\frac{2}{1-sin^2A}

\frac{2}{cos^2A}

= 2 sec2 A

= R.H.S.

Hence proved.

Question 26. \frac{1+sinθ}{cosθ}+\frac{cosθ}{1+sinθ}=2secθ    

Solution:

We have,

L.H.S. = \frac{1+sinθ}{cosθ}+\frac{cosθ}{1+sinθ}



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

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

\frac{2(1+sinθ)}{cosθ(1+sinθ)}

\frac{2}{cosθ}

= 2 sec θ

= R.H.S.

Hence proved.

Question 27. \frac{(1+sinθ)^2+(1-sinθ)^2}{2cos^2θ}=\frac{1+sin^2θ}{1-sin^2θ}

Solution:

We have,

L.H.S. = \frac{(1+sinθ)^2+(1-sinθ)^2}{2cos^2θ}

\frac{1+sin^2θ+2sinθ+1+sin^2θ-2sinθ}{2cos^2θ}

\frac{2(1+sin^2θ)}{2(1-sin^2θ)}

\frac{1+sin^2θ}{1-sin^2θ}

= R.H.S.

Hence proved.

Question 28. \frac{1+tan^2θ}{1+cot^2θ}=tan^2θ

Solution:

We have,

L.H.S. = \frac{1+tan^2θ}{1+cot^2θ}

= sec2 θ/cosec2 θ

\frac{1}{cos^2θ}×\frac{sin^2θ}{1}

= tan2 θ

= R.H.S.

Hence proved.

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