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

Question

asked Sep 22, 2020 170k views

1 Answer

Shamseer PC · Answer 1 · 2020-09-27T07:45:47+0000

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.