Question

Prove the following trigonometric identities.

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

Answer 1

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

Here, we will first solve the LHS.

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

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

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

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

Further, multiplying both numerator and denominator by 1 + sin A we get

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

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

`= (1 s sin^2 A)/(1 + sin A)^2`

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

So,

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

Hence proved