Question

Show that? `(1-tan^2x)/(1+tan^2x) = cos^2-sin^2x`

Answer 1

Consider the LHS of the given expression:

`LHS = (1-tan^2x)/(1+tan^2x)`

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

`\ \ \ \ \ \ \ \ = ((cos^2-sin^2x)/cos^2x)/((cos^2x+sin^2x)/cos^2x)`

`\ \ \ \ \ \ \ \ = ((cos^2-sin^2x)/cos^2x) * (cos^2x/(cos^2x+sin^2x))`

`\ \ \ \ \ \ \ \ = (cos^2-sin^2x)/(cos^2x+sin^2x)`

`\ \ \ \ \ \ \ \ = (cos^2-sin^2x)/1 \ \ \ ` , as `cos^2A+sin^2xA -= 1`

`\ \ \ \ \ \ \ \ = cos^2-sin^2x \ \ \ ` QED