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

Last Updated : 30 Apr, 2021

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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

= (sin²θ/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 + tan²A) = 1

Solution:

We have,

L.H.S. = sin² A + 1/(1 + tan² A)

= sin² A + 1/(sec² A)

= sin² A + cos²A

= 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) (cos²θ)

= 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.