Question

Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`

Answer 1

We need to prove `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`

Now, rationalising the L.H.S, we get

`(1 - cos A)/(1 + cos A) = ((1 - cos A)/(1 +  cos A)) ((1 - cos A)/(1 - cos A))`

`= (1 - cos A)^2/(1 - cos^2 A)`      (using `a^2 - b^2 = (a + b)(a - b))`

` = (1 + cos^2 A - 2 cos A)/sin^2 A`    (Using `sin^2 theta = 1 - cos^2 theta`)

`= 1/sin^2 A + cos^2 A/sin^2 A - (2 cos A)/sin^2 A`

Using `cosec theta = 1/sin theta` and `cot theta = cos theta/sin theta` we get

`1/sin^2 A + cos^2 A/sin^2 A - (2 cos A)/sin^2 A = cosec^2 A + cot^2 A - 2 cot A cosec A`

` (cot A - cosec A)^2`    (Using `(a + b)^2 = a^2 + b^2 + 2ab`)

Hence proved.

Answer 2

LHS = `(1 - cos θ)/(1 + cos θ)`

= `(1 - cos θ)/(1 + cos θ) xx (1 - cos θ)/(1 - cos θ)`

= `(1 - cos θ)^2/(1 - cos^2 θ)`

= `(1 - cos θ)^2/(sin^2 θ)`

= `[(1 - cosθ)/(sin θ)]^2`

= `[ 1/sinθ  - cosθ/sin θ ]^2`

= ( cosec θ - cot θ )²

= [ - (cot θ - cosec θ)]²

= (cot θ - cosec θ)²

= RHS

Hence proved.