# Prove that `1 - 2*cos^2x = (tan^2x - 1)/(tan^2x + 1)`

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We know;

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

`cos2x = 2cos^2x-1 = cos^2x-sin^2x`

`2cos^2x = cos2x+1`

Start from LHS

`1-2cos^2x`

= `1-(cos2x+1)`

= `-cos2x`

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

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

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

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

**Therefore** `1-2cos^2x = (tan^2x-1)/(tan^2x+1)`

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

Start from the right hand side

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

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

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

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

=> `1 - cos^2x - cos^2x`

=> `1 - 2*cos^2x`

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