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

• Last Updated : 30 Apr, 2021

### 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

= (sin2 θ/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

= 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. 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 + tan2 A) = 1

Solution:

We have,

L.H.S. = sin2 A + 1/(1 + tan2 A)

= sin2 A + 1/(sec2 A)

= sin2 A + cos2 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

Solution:

We have,

L.H.S. = (1 – sin θ)/(1 + sin θ)

= (sec θ – tan θ)2

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= 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)

= 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)

= 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 + 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) (cos2 θ)

= 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.

### (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. (cos2 θ/sin θ) – cosec θ + sin θ = 0

Solution:

We have,

L.H.S. = (cos2 θ/sin θ) – cosec θ + sin θ

= 0

= 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. =

= 2 sec θ

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

### Question 28.

Solution:

We have,

L.H.S. =

= sec2 θ/cosec2 θ

= tan2 θ

= R.H.S.

Hence proved.

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