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\[\tan^2x-\tan^2x\sin^2x=\tan^2x(1-\sin^2x)\]and what is the identity for \(1-\sin^2x\) ?

cos^2x

right, so now what do you have on the left?

tan^2x-tan^2xsin^2x=tan^2x

no, I mean after what we just did...

tan^2x

\[\tan^2x-\tan^2x\sin^2x=\tan^2x(1-\sin^2x)=\tan^2x\cos^2x=?\]

use\[\tan x=\frac{\sin x}{\cos x}\]

that would leave you with sin^2x as the answer