Question

`Prove the following trigonometric identities.

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

Answer 1

We need to prove  `(sec A - tan A)^2  = (1 - sin A)/(1 +  sin A)`

Here, we will first solve the L.H.S.

Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta` we get

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

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

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

Further using the property  `sin^2 theta + cos^2 theta = 1` we get

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

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

`= (1 - sin A)/(1 +  sin A)`

henc e proved