(1/cot^2 50 degrees) - (1/cos^2 50 degrees) = ?
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$\displaystyle \cos^2 x +sin^2 x \equiv 1$ $\displaystyle 1+\tan^2 x \equiv \sec^2 x$ $\displaystyle \tan^2 x - \sec ^2 x \equiv ...$
Originally Posted by avatartiger2 (1/cot^2 50 degrees) - (1/cos^2 50 degrees) = ? $\displaystyle \frac{1}{\text{Cot}(\alpha )^2}-\frac{1}{\text{Cos}(\alpha )^2}=-1$
Last edited by MaxJasper; Sep 12th 2012 at 04:25 PM.
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