# Class 10 RD Sharma Solutions – Chapter 6 Trigonometric Identities – Exercise 6.1 | Set 2

**Prove the following trigonometric identities:**

**Question 29. **** **

**Solution:**

We have,

L.H.S. =

=

=

= 1 + cos θ

=

=

=

= R.H.S.

Hence proved.

**Question 30. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

= 1 + tanθ + cotθ

= R.H.S.

Hence proved.

**Question 31. sec**^{6} θ = tan^{6} θ + 3 tan^{2} θ sec^{2} θ + 1

^{6}θ = tan

^{6}θ + 3 tan

^{2}θ sec

^{2}θ + 1

**Solution: **

We know,

sec

^{2}θ − tan^{2}θ = 1On cubing both sides, we get,

=> (sec

^{2}θ − tan^{2}θ)^{3}= 1=> sec

^{6}θ − tan^{6}θ − 3sec^{2}θ tan^{2}θ(sec^{2}θ − tan^{2}θ) = 1=> sec

^{6}θ − tan^{6}θ − 3sec^{2}θ tan^{2}θ = 1=> sec

^{6}θ = tan^{6}θ + 3sec^{2}θ tan^{2}θ + 1

Hence proved.

**Question 32. cosec**^{6} θ = cot^{6} θ + 3cot^{2} θ cosec^{2} θ + 1

^{6}θ = cot

^{6}θ + 3cot

^{2}θ cosec

^{2}θ + 1

**Solution:**

We know,

cosec

^{2}θ − cot^{2}θ = 1On cubing both sides,

=> (cosec

^{2}θ − cot^{2}θ)^{3}= 1=> cosec

^{6}θ − cot^{6}θ − 3cosec^{2}θ cot^{2}θ (cosec^{2}θ − cot^{2}θ) = 1=> cosec

^{6}θ − cot^{6}θ − 3cosec^{2}θ cot^{2}θ = 1=> cosec

^{6 }θ = cot^{6}θ + 3 cosec^{2}θ cot^{2}θ + 1

Hence proved.

**Question 33. **

**Solution:**

We have,

L.H.S. =

=

=

= sin θ/cos θ

= tan θ

= R.H.S.

Hence proved.

**Question 34. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= R.H.S.

Hence proved.

**Question 35. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

= R.H.S.

Hence proved.

**Question 36. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= R.H.S.

Hence proved.

**Question 37. (i) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= sec A + tan A

= R.H.S.

Hence proved.

**(ii) **

**Solution:**

We have,

L.H.S. =

=

=

= 2 cosec A

= R.H.S.

Hence proved.

**Question 38. (i) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= 2 cosec θ

= R.H.S.

Hence proved.

**(ii) **

**Solution:**

We have,

L.H.S. =

=

=

= 2 sec θ

= R.H.S.

Hence proved.

**(iii) **** **

**Solution:**

We have,

L.H.S. =

=

= 2 cosec θ

= R.H.S.

Hence proved.

**(iv) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= R.H.S.

Hence proved.

**Question 39. (sec A – tan A)**^{2} =

^{2}=

**Solution:**

We have,

L.H.S. = (sec A – tan A)

^{2}=

=

=

=

=

= R.H.S.

Hence proved.

**Question 40. **

**Solution:**

We have,

L.H.S. =

=

=

=

= (cosec A – cot A)

^{2}= (cot A – cosec A)

^{2}= R.H.S.

Hence proved.

**Question 41. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

= 2 cosec A cot A

= R.H.S.

Hence proved.

**Question 42. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

= sin A + cos A

= R.H.S.

Hence proved.

**Question 43. **** **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= 2 sec

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

Hence proved.

**Question 44. **

**Solution:**

We have,

L.H.S. =

=

=

=

= 1

= R.H.S.

Hence proved.

**Question 45. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

= R.H.S.

Hence proved.

**Question 46. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= cos θ/sin θ

= cot θ

= R.H.S.

Hence proved.

**Question 47. (i) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

= sec θ + tan θ

= 1/cos θ + sin θ/cos θ

=

= R.H.S.

Hence proved.

**(ii) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

=

= R.H.S.

Hence proved.

**Question 48.**

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

=

=

= cosec θ + cot θ

= R.H.S.

Hence proved.

**Question 49. (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ**

**Solution:**

We have,

L.H.S. = (sin θ + cos θ) (tan θ + cot θ)

= sin

^{2}θ/cosθ + cos θ + sin θ + cos^{2}θ/sin θ= sin θ (1 + tan θ) + (cos θ/tan θ) (1 + tan θ)

= (1 + tan θ) (sin θ + cos θ/tan θ)

=

=

= sec θ + cosec θ

= R.H.S.

Hence proved.

**Question 50. **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

= 2/sin A

= 2 cosec A

= R.H.S.

Hence proved.

**Question 51. 1 + **** = cosec θ**

**Solution:**

We have,

L.H.S. = 1 +

= 1 +

= 1+

= 1 + cosec θ − 1

= cosec θ

= R.H.S.

Hence proved.

**Question 52.**

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

= 2 sin θ/cos θ

= 2 tan θ

= R.H.S.

Hence proved.

**Question 53. (1 + tan**^{2} A) + (1 + 1/tan^{2} A) = 1/(sin^{2} A − sin^{4} A)

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

^{2}A) = 1/(sin

^{2}A − sin

^{4}A)

**Solution:**

We have,

L.H.S. = (1 + tan

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

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

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

=

= 1/(sin

^{2}A − sin^{4}A)= R.H.S.

Hence proved.

**Question 54. sin**^{2} A cos^{2} B −** cos**^{2} A sin^{2} B = sin^{2} A − **sin**^{2} B

^{2}A cos

^{2}B

^{2}A sin

^{2}B = sin

^{2}A

^{2}B

**Solution:**

We have,

L.H.S. = sin

^{2}A cos^{2}B − cos^{2}A sin^{2}B= sin

^{2}A (1 − sin^{2}B) − sin^{2}B (1 − sin^{2}A)= sin

^{2}A− sin^{2}A sin^{2}B − sin^{2}B + sin^{2}A sin^{2}B= sin

^{2}A − sin^{2}B= R.H.S.

Hence Proved.

**Question 55. (i) **** **

**Solution:**

We have,

L.H.S. =

=

=

=

= cot A tan B

= R.H.S.

Hence proved.

**(ii)**

**Solution:**

We have,

L.H.S. =

=

=

=

= tan A tan B

= R.H.S.

Hence proved.

**Question 56. cot**^{2 }A cosec^{2 }B − cot^{2 }B cosec^{2 }A = cot^{2 }A − cot^{2}B

^{2 }A cosec

^{2 }B − cot

^{2 }B cosec

^{2 }A = cot

^{2 }A − cot

^{2}B

**Solution:**

We have,

L.H.S. = cot

^{2 }A cosec^{2 }B − cot^{2}B cosec^{2 }A= cot

^{2}A (1 + cot^{2}B) − cot^{2}B (1 + cot^{2}A)= cot

^{2 }A + cot^{2 }A cot^{2 }B − cot^{2 }B − cot^{2 }B cot^{2 }A= cot

^{2 }A − cot^{2 }B= R.H.S.

Hence proved.

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