Recall the Pythagorean identity,
[tex]1+\tan^2x=\sec^2x[/tex]
Then
[tex]\sec^2x-2\tan^2x=\sec^2x-2(\sec^2x-1)=2-\sec^2x[/tex]
Recall the double angle identity for cosine:
[tex]\cos(2x)=2\cos^2x-1[/tex]
from which we get
[tex]2=\dfrac{\cos(2x)+1}{\cos^2x}=(\cos(2x)+1)\sec^2x[/tex]
Substituting this above gives
[tex]\sec^2x-2\tan^2x=(\cos(2x)+1)\sec^2x-\sec^2x=\sec^2x(\cos(2x)+1-1)[/tex]
[tex]\implies\sec^2x-2\tan^2x=\sec^2x\cos(2x)[/tex]
as required.
