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

• Last Updated : 30 Apr, 2021

Solution:

We have,

L.H.S. =

= 1 + cos Î¸

= R.H.S.

Hence proved.

### Question 30.

Solution:

We have,

L.H.S. =

= 1 + tanÎ¸ + cotÎ¸

= R.H.S.

Hence proved.

### Question 31. sec6 Î¸ = tan6 Î¸ + 3 tan2 Î¸ sec2 Î¸ + 1

Solution:

We know,

sec2 Î¸ âˆ’ tan2 Î¸ = 1

On cubing both sides, we get,

=> (sec2Î¸ âˆ’ tan2Î¸)3 = 1

=> sec6 Î¸ âˆ’ tan6 Î¸ âˆ’ 3sec2 Î¸ tan2 Î¸(sec2 Î¸ âˆ’ tan2 Î¸) = 1

=> sec6 Î¸ âˆ’ tan6 Î¸ âˆ’ 3sec2 Î¸ tan2 Î¸ = 1

=> sec6 Î¸ = tan6 Î¸ + 3sec2 Î¸ tan2 Î¸ + 1

Hence proved.

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

Solution:

We know,

cosec2 Î¸ âˆ’ cot2 Î¸ = 1

On cubing both sides,

=> (cosec2 Î¸ âˆ’ cot2 Î¸)3 = 1

=> cosec6 Î¸ âˆ’ cot6 Î¸ âˆ’ 3cosec2 Î¸ cot2 Î¸ (cosec2 Î¸ âˆ’ cot2 Î¸) = 1

=> cosec6 Î¸ âˆ’ cot6 Î¸ âˆ’ 3cosec2 Î¸ cot2 Î¸ = 1

=> cosec6 Î¸ = cot6 Î¸ + 3 cosec2 Î¸ cot2 Î¸ + 1

Hence proved.

Solution:

We have,

L.H.S. =

= sin Î¸/cos Î¸

= tan Î¸

= 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 Î¸

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 2 sec Î¸

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 2 cosec Î¸

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

### Question 39. (sec A â€“ tan A)2 =

Solution:

We have,

L.H.S. = (sec A â€“ tan A)2

= R.H.S.

Hence proved.

### Question 40.

Solution:

We have,

L.H.S. =

= (cosec A â€“ cot A)2

= (cot A â€“ 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 Î¸/sin Î¸

= cot Î¸

= R.H.S.

Hence proved.

### Question 47. (i)

Solution:

We have,

L.H.S. =

= sec Î¸ + tan Î¸

= 1/cos Î¸ + sin Î¸/cos Î¸

= 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 Î¸ + cot Î¸

= R.H.S.

Hence proved.

### Question 49. (sin Î¸ + cos Î¸) (tan Î¸ + cot Î¸) = sec Î¸ + cosec Î¸

Solution:

We have,

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

= sin2 Î¸/cosÎ¸ + cos Î¸ + sin Î¸ + cos2 Î¸/sin Î¸

= sin Î¸ (1 + tan Î¸) + (cos Î¸/tan Î¸) (1 + tan Î¸)

= (1 + tan Î¸) (sin Î¸ + cos Î¸/tan Î¸)

= sec Î¸ + cosec Î¸

= 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 Î¸

Solution:

We have,

L.H.S. = 1 +

= 1 +

= 1+

= 1 + cosec Î¸ âˆ’ 1

= cosec Î¸

= R.H.S.

Hence proved.

### Question 52.

Solution:

We have,

L.H.S. =

= 2 sin Î¸/cos Î¸

= 2 tan Î¸

= R.H.S.

Hence proved.

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

= R.H.S.

Hence proved.

### Question 54. sin2 A cos2 B âˆ’ cos2 A sin2 B = sin2 A âˆ’ sin2 B

Solution:

We have,

L.H.S. = sin2 A cos2 B âˆ’ cos2 A sin2 B

= sin2 A (1 âˆ’ sin2 B) âˆ’ sin2 B (1 âˆ’ sin2 A)

= sin2 Aâˆ’ sin2 A sin2 B âˆ’ sin2 B + sin2 A sin2 B

= sin2A âˆ’ 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 âˆ’ cot2 B cosec2 A = cot2 A âˆ’ cot2B

Solution:

We have,

L.H.S. = cot2 A cosec2 B âˆ’ cot2 B cosec2 A

= cot2 A (1 + cot2 B) âˆ’ cot2 B (1 + cot2 A)

= cot2 A + cot2 A cot2 B âˆ’ cot2 B âˆ’ cot2 B cot2 A

= cot2 A âˆ’ cot2 B

= R.H.S.

Hence proved.

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