# 10th 7th chapter solution for trigonometry part1

This page 10th 7th chapter solution for trigonometry  part1 is going to provide you solution for every problems that you find in the exercise no 7.1

## 10th 7th chapter solution for trigonometry part1

(1) Determine whether each of the following is an identity or not.

(i) cos²θ + sec²θ = 2 + sin θ

Solution:

To check whether the given question is an identity or not,first  we have to simplify the left side question. So that we have to get the right side answer.

L.H.S

= cos²θ + sec²θ

= 1 - sin²θ + 1 + tan²θ

= 2 + tan²θ - sin²θ

The answer that we got didn't match the right side answer. So this is not an identity.

(ii) cot²θ + cos θ = sin² θ

L.H.S

= cot²θ + cos θ

= cosec²θ - 1 + cos θ

This does not match the with the right side answer. So this is not an identity.

(2) Prove the following identities

(i) sec²θ + cosec²θ = sec²θ  cosec²θ

L.H.S

= sec²θ + cosec²θ

= (1/cos²θ)+ (1/sin²θ)

taking L.C.M

= (sin²θ + cos²θ)/(cos²θ sin²θ)

value of sin²θ + cos²θ = 1

= 1/(cos²θ sin²θ)

= (1/cos²θ)(1/sin²θ)

= sec²θ cosec²θ

In the page 10th 7th chapter solution for trigonometry part1 we are going to see the solution of next problem

(ii)  sin θ /(1-cos θ) = cosec θ + cot θ

L.H.S

= sin θ /(1-cos θ)

multiplying by the conjugate of denominator we get,

= [sin θ /(1-cos θ)] x [(1+cos θ)/(1+cos θ)]

instead of (1+cos θ)/(1+cos θ) we can write 1 - cos²θby using the algebraic formula

= [sin θ (1 + cosθ)]/(1-cos²θ)

= [sin θ (1 + cosθ)]/sin²θ

= (1 + cosθ)]/sinθ

= (1/sinθ) + (cosθ/sinθ)

= cosec θ + cot θ

(iii) (1-sin θ)/(1+sin θ) = sec θ - tan θ

L.H.S

= √(1-sin θ)/(1+sin θ)

= (1-sin θ)/(1+sin θ) x (1-sin θ)/(1-sin θ)

= (1-sin θ)²/[(1+sin θ) x (1-sin θ)]

= (1-sin θ)²/(1²- sin²θ)

= (1-sin θ)²/(cos²θ)

= [(1-sin θ)/(cosθ)]²

= [(1-sin θ)/(cosθ)]

= [(1/cosθ)-(sin θ/cosθ)]

= sec θ - tan θ

= R.H.S