Question

Prove that sec^2 x - 2 tan^2 x = sec^2 x cos 2x.​

Answer 1

Recall the Pythagorean identity,

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

Then

`\sec^2x-2\tan^2x=\sec^2x-2(\sec^2x-1)=2-\sec^2x`

Recall the double angle identity for cosine:

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

from which we get

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

Substituting this above gives

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

`\implies\sec^2x-2\tan^2x=\sec^2x\cos(2x)`

as required.