Question

What is `tan^2theta/(1-sin^2theta` in terms of non-exponential trigonometric functions?

Answer 1

Please see the explanation below

Explanation:

We need

`cos^2x+sin^2x=1`

`tan^2x+1=sec^2x`

Therefore,

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

`=((secx+1)(secx-1))/((1+sinx)(1-sinx))`