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

**Prove the following trigonometric identities:**

**Question 1. (1 – cos**^{2} A) cosec^{2} A = 1

^{2}A) cosec

^{2}A = 1

**Solution:**

We have,

L.H.S. = (1 – cos

^{2}A) cosec^{2}ABy using the identity, sin

^{2}A + cos^{2}A = 1, we get,= (sin

^{2}A) (cosec^{2}A)= sin

^{2}A × (1/sin^{2}A)= 1

= R.H.S.

Hence proved.

**Question 2. (1 + cot**^{2} A) sin^{2} A = 1

^{2}A) sin

^{2}A = 1

**Solution: **

We have,

L.H.S. = (1 + cot

^{2}A) sin^{2}ABy using the identity, cosec

^{2}A = 1 + cot^{2}A, we get,= cosec

^{2}A sin^{2}A= (1/sin

^{2}A) × sin^{2}A= 1

= R.H.S

Hence proved.

**Question 3. tan**^{2} θ cos^{2} θ = 1 − cos^{2} θ

^{2}θ cos

^{2}θ = 1 − cos

^{2}θ

**Solution: **

We have,

L.H.S. = tan

^{2}θ cos^{2}θ= (sin

^{2 }θ/cos^{2}θ) (cos^{2}θ)= sin

^{2}θ= 1 − cos

^{2}θ= R.H.S.

Hence proved.

**Question 4. cosec θ √(1 – cos**^{2} θ) = 1

^{2}θ) = 1

**Solution:**

We have,

L.H.S. = cosec θ √(1 – cos

^{2}θ)= cosec θ √(sin

^{2}θ)= cosec θ sin θ

= (1/sin θ) (sin θ)

= 1

= R.H.S.

Hence proved.

**Question 5. (sec**^{2} θ − 1)(cosec^{2} θ − 1) = 1

^{2}θ − 1)(cosec

^{2}θ − 1) = 1

**Solution:**

We have,

L.H.S. = (sec

^{2}θ − 1)(cosec^{2}θ − 1)By using the identities sec

^{2}θ − tan^{2}θ = 1 and cosec^{2}θ − cot^{2}θ = 1, we have,= tan

^{2}θ cot^{2}θ= (tan

^{2}θ) (1/tan^{2}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

= (tan

^{2}θ + 1)/tan θ= sec

^{2}θ/tan 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)

=

=

=

= (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)

=

=

=

= (1 – sin θ)/cos θ

= R.H.S.

Hence proved.

**Question 9. cos**^{2} θ + 1/(1 + cot^{2} θ) = 1

^{2}θ + 1/(1 + cot

^{2}θ) = 1

**Solution:**

We have,

L.H.S. = cos

^{2}θ + 1/(1 + cot^{2}θ)= cos

^{2}θ + 1/(cosec^{2}θ)= cos

^{2}θ + sin^{2}θ= 1

= R.H.S.

Hence proved.

**Question 10. sin**^{2} A + 1/(1 + tan^{2 }A) = 1

^{2}A + 1/(1 + tan

^{2 }A) = 1

**Solution:**

We have,

L.H.S. = sin

^{2}A + 1/(1 + tan^{2}A)= sin

^{2}A + 1/(sec^{2}A)= sin

^{2}A + cos^{2 }A= 1

= R.H.S.

Hence proved.

**Question 11. **** = cosec θ − cot θ**

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

= 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

=

=

=

= 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)

=

=

=

=

=

= cosec θ + cot θ

= R.H.S.

Hence proved.

**Question 14. (1 – sin θ)/(1 + sin θ) = (sec θ – tan θ)**^{2}

^{2}

**Solution:**

We have,

L.H.S. = (1 – sin θ)/(1 + sin θ)

=

=

=

=

=

= (sec θ – tan θ)

^{2}= R.H.S.

Hence proved.

**Question 15. **

**Solution:**

We have,

L.H.S. =

=

=

= cos θ/sin θ

= cot θ

= R.H.S.

Hence proved.

**Question 16. tan**^{2} θ − sin^{2} θ = tan^{2} θ sin^{2} θ

^{2}θ − sin

^{2}θ = tan

^{2}θ sin

^{2}θ

**Solution:**

We have,

L.H.S. = tan

^{2}θ − sin^{2}θ= sin

^{2}θ/cos^{2}θ − sin^{2}θ= sin

^{2}θ(1/cos^{2}θ − 1)=

= sin

^{2}θ (sin^{2}θ/cos^{2}θ)= tan

^{2}θ sin^{2}θ= R.H.S.

Hence proved.

**Question 17. (cosec θ + sin θ)(cosec θ – sin θ) = cot**^{2}θ + cos^{2}θ

^{2}θ + cos

^{2}θ

**Solution:**

We have,

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

= cosec

^{2}θ – sin^{2}θ= (1 + cot

^{2}θ) – (1 – cos^{2}θ)= 1 + cot

^{2}θ – 1 + cos^{2}θ= cot

^{2}θ + cos^{2}θ= R.H.S.

Hence proved.

**Question 18. (sec θ + cos θ) (sec θ – cos θ) = tan**^{2} θ + sin^{2} θ

^{2}θ + sin

^{2}θ

**Solution:**

We have,

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

= sec

^{2}θ – cos^{2}θ= (1 + tan

^{2}θ) – (1 – sin^{2}θ)= 1 + tan

^{2}θ – 1 + sin^{2}θ= tan

^{2}θ + sin^{2}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)

=

=

=

= 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)

=

=

=

= 1

= R.H.S

Hence proved.

**Question 21. (1 + tan**^{2} θ)(1 – sin θ)(1 + sin θ) = 1

^{2}θ)(1 – sin θ)(1 + sin θ) = 1

**Solution:**

We have,

L.H.S. = (1 + tan

^{2}θ)(1 – sin θ)(1 + sin θ)= (sec

^{2}θ) (1 – sin^{2}θ)= (sec

^{2}θ) (cos^{2 }θ)= 1

= R.H.S

Hence proved.

**Question 22. sin**^{2} A cot^{2} A + cos^{2} A tan^{2} A = 1

^{2}A cot

^{2}A + cos

^{2}A tan

^{2}A = 1

**Solution:**

We have,

L.H.S. = sin

^{2}A cot^{2}A + cos^{2}A tan^{2}A= sin

^{2}A (cos^{2}A/sin^{2}A) + cos^{2}A (sin^{2}A/cos^{2}A)= cos

^{2}A + sin^{2}A= 1

= R.H.S.

Hence proved.

**Question 23. **

**(i) cot θ – tan θ =**

**Solution:**

We have,

L.H.S. = cot θ – tan θ

= cos θ/sin θ – sin θ/cos θ

=

=

=

=

= R.H.S.

Hence proved.

**(ii) tan θ – cot θ** **=**

**Solution:**

We have,

L.H.S. = tan θ – cot θ

= sin θ/cos θ – cos θ/sin θ

=

=

=

=

= R.H.S.

Hence proved.

**Question 24. (cos**^{2} θ/sin θ) – cosec θ + sin θ = 0

^{2}θ/sin θ) – cosec θ + sin θ = 0

**Solution:**

We have,

L.H.S. = (cos

^{2}θ/sin θ) – cosec θ + sin θ=

=

=

= 0

= R.H.S.

Hence proved.

**Question 25. **

**Solution:**

We have,

L.H.S. =

=

=

=

= 2 sec

^{2}A= R.H.S.

Hence proved.

**Question 26. **** **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= 2 sec θ

= R.H.S.

Hence proved.

**Question 27.**

**Solution:**

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

**Question 28. **

**Solution:**

We have,

L.H.S. =

= sec

^{2}θ/cosec^{2}θ=

= tan

^{2}θ= R.H.S.

Hence proved.

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