Please take a look at the attached question, thank you!

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- Oct 15th 2008, 07:32 AMgearshifterSimplify Trig
Please take a look at the attached question, thank you!

- Oct 15th 2008, 08:26 AMChop Suey
Recall that:

$\displaystyle \tan^2{x} = \sec^2{x} - 1$

Notice that $\displaystyle 1-\sec^2{x} = -(\sec^2{x}-1)$

You end up with $\displaystyle -\cot{x}\sin{x}$...Remember that $\displaystyle \cot{x} = \frac{\cos{x}}{\sin{x}}$