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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.

$$ \tan^2x - \sec^2x $$

$$ (\sin x / \cos x)^2 - (x / \cos x)^2 $$

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    $\begingroup$ The problem is that $\sec(x)=\dfrac{1}{\cos(x)}\neq\dfrac{x}{\cos(x)}.$ $\endgroup$ – ՃՃՃ Jan 17 '13 at 6:40
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$$\tan^2x-\sec^2x=\frac{\sin^2x-1}{\cos^2x}=\frac{-\cos^2x}{\cos^2x}=-1.$$

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  • $\begingroup$ where did the sin^2x go $\endgroup$ – sam Jan 17 '13 at 6:43
  • $\begingroup$ $\cos^2x+\sin^2x=1$ $\endgroup$ – Jonathan Jan 17 '13 at 6:44
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$$ \tan^2x - \sec^2x $$ since : $\sec ^2x=1+\tan ^2 x$

so $$\tan^2x - (1+\tan ^2x)\implies -1$$

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