The trigonometric identity `2*cos^2x - 1 = (1 - tan^2)/(1 + tan^2 x)` has to be proved.

Let's start from the right hand side.

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

Substitute `tan x = sin x/cos x`

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

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

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

It is known that `cos^2x + sin^2x = 1`

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

= `2*cos^2x - cos^2x - sin^2x`

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

= `2*cos^2x - 1`

**This proves that **`2*cos^2x - 1 = (1 - tan^2x)/(1 + tan^2 x)`

To prove the trigonometric identity: `2cos^2(x) -1 = (1-tan^2(x))/(1+tan^2(x))` ,

Substitute: `tan^2= sec^2 -1` and

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

Note that `-(sec^2(x)-1) = -sec^2(x) +1`

`=(2 -sec^2 (x))/(sec^2(x))`

Express into two fractions:

`= 2/(sec^2(x)) - (sec^2(x))/(sec^2(x))`

Simplify and substitute `cos^2(x) = 1/(sec^2(x))`

`= 2cos^2(x) -1`

This proves that `2cos^2(x)-1 = (1-tan^2(x))/(1+tan^2(x))`