Question

Prove the following trigonometric identities:

`sqrt((1 - cos A)/(1 + cos A)) + sqrt((1 + cos A)/(1 - cos A)) = 2  "cosec"  A`

Answer 1

Given: `S = sqrt((1 - cos A)/(1 + cos A)) + sqrt((1 + cos A)/(1 - cos A))`, with A such that the square roots are defined (cos A ≠ ±1, i.e. A ≠ nπ).

To Prove: S = 2 cosec A.

Proof [Step-wise]:

1. Let `t = sqrt((1 - cos A)/(1 + cos A))`.

Then the second term is `1/t`.

So `S = t + 1/t`.

2. Compute `t + 1/t = (t^2 + 1)/t`.

3. Compute t² + 1:

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

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

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

= `2/(1 + cos A)`

4. Therefore `S = (2/(1 + cos A))/t` 

= `2/(t(1 + cos A))`

Substitute `t = sqrt((1 - cos A)/(1 + cos A))`: 

`S = 2/sqrt((1 - cos A)(1 + cos A))` 

= `2/sqrt(1 - cos^2 A)`

= `2/|sin A|`

5. If we assume sin A > 0 so |sin A| = sin A, this becomes S = `2/sin A` = 2 cosec A.

For A with sin A > 0 and cos A ≠ ±1, `sqrt((1 - cos A)/(1 + cos A)) + sqrt((1 + cos A)/(1 - cos A)) = 2  "cosec"  A`. More generally, for real A with cos A ≠ ±1 the sum equals `2/|sin A|`.