I am trying to express this problem in terms of sin/cos and simplify. I couldn't figure out where to go, I tried as best I could. I know the answer is -1 but I am more interested to know how to do this problem.
$$ \tan^2x - \sec^2x $$
$$ (\sin x / \cos x)^2 - (x / \cos x)^2 $$
$\endgroup$ 12 Answers
$\begingroup$$$\tan^2x-\sec^2x=\frac{\sin^2x-1}{\cos^2x}=\frac{-\cos^2x}{\cos^2x}=-1.$$
$\endgroup$ 2 $\begingroup$$$ \tan^2x - \sec^2x $$ since : $\sec ^2x=1+\tan ^2 x$
so $$\tan^2x - (1+\tan ^2x)\implies -1$$
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