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• RD Sharma Class 10 Solutions

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

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. sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1

Solution:

We know,

sec2 θ − tan2 θ = 1

On cubing both sides, we get,

=> (sec2θ − tan2θ)3 = 1

=> sec6 θ − tan6 θ − 3sec2 θ tan2 θ(sec2 θ − tan2 θ) = 1

=> sec6 θ − tan6 θ − 3sec2 θ tan2 θ = 1

=> sec6 θ = tan6 θ + 3sec2 θ tan2 θ + 1

Hence proved.

### Question 32. cosec6 θ = cot6 θ + 3cot2 θ cosec2 θ + 1

Solution:

We know,

cosec2 θ − cot2 θ = 1

On cubing both sides,

=> (cosec2 θ − cot2 θ)3 = 1

=> cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ (cosec2 θ − cot2 θ) = 1

=> cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ = 1

=> cosec6 θ = cot6 θ + 3 cosec2 θ cot2 θ + 1

Hence proved.

Solution:

We have,

L.H.S. =

= sin θ/cos θ

= tan θ

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= sec A + tan A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 2 cosec A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 2 cosec θ

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 2 sec θ

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 2 cosec θ

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

### Question 39. (sec A – tan A)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.

Solution:

We have,

L.H.S. =

= sin A + cos A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 2 sec2 A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 1

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

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.

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)

= sin2 θ/cosθ + cos θ + sin θ + cos2 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.

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.

Solution:

We have,

L.H.S. =

= 2 sin θ/cos θ

= 2 tan θ

= R.H.S.

Hence proved.

### Question 53. (1 + tan2 A) + (1 + 1/tan2 A) = 1/(sin2 A − sin4 A)

Solution:

We have,

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

= (1 + sin2 A/cos2 A) + (1 + cos2 A/sin2 A)

= 1/cos2 A + 1/sin2 A

= 1/(sin2 A − sin4 A)

= R.H.S.

Hence proved.

### Question 54. sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B

Solution:

We have,

L.H.S. = sin2 A cos2 B − cos2 A sin2 B

= sin2 A (1 − sin2 B) − sin2 B (1 − sin2 A)

= sin2 A− sin2 A sin2 B − sin2 B + sin2 A sin2 B

= sin2A − sin2 B

= R.H.S.

Hence Proved.

Solution:

We have,

L.H.S. =

= cot A tan B

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= tan A tan B

= R.H.S.

Hence proved.

### Question 56. cot2 A cosec2 B − cot2 B cosec2 A = cot2 A − cot2B

Solution:

We have,

L.H.S. = cot2 A cosec2 B − cot2 B cosec2 A

= cot2 A (1 + cot2 B) − cot2 B (1 + cot2 A)

= cot2 A + cot2 A cot2 B − cot2 B − cot2 B cot2 A

= cot2 A − cot2 B

= R.H.S.

Hence proved.

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