Prove that `(1-tan^2 theta) / (1+tan^2 theta) -= cos(2 theta)`?

We seek to prove that:

`(1-tan^2 theta) / (1+tan^2 theta) -= cos(2 theta)`

We will require the following trigonometric definitions/identities:

`tan phi = sin phi / cos phi`
`sin^2 phi + cos^phi -=1`
`cos 2phi -= cos^2 phi - sin^2 phi`

Consider the LHS of the given expression:

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

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

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

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

`\ \ \ \ \ \ \ \ = (cos 2theta)/(1)`

`\ \ \ \ \ \ \ \ = cos 2theta`

`\ \ \ \ \ \ \ \ = RHS \ \ \ \ ` QED

Here are the identities that I used:

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

`tan^2theta=sin^2theta/cos^2theta`

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

`cos(2theta)=cos^2theta-sin^2theta`

Now here's the actual proof:

`(1-tan^2 theta) / (1+tan^2 theta)=cos(2 theta)`

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

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

`(-sin^2theta+cos^2theta)/cancel(sin^2theta+cos^2theta)=cos(2theta)`

`-sin^2theta+cos^2theta=cos(2theta)`

`cos^2theta-sin^2theta=cos(2theta)`

`cos(2theta)=cos(2theta)`

`1 + tan^2 theta = sec^2 theta = 1/cos^2 theta`

`1 - 2sin^2 theta = 2cos^2 theta - 1 = cos^2 theta - sin^2 theta = cos (2theta)`

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

`((cos^2theta-sin^2theta)/cancel(cos^2(theta))/ 1/cancel(cos^2 (theta))) = cos^2(theta) - sin^2theta = cos(2theta)`