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I don't need the answer, I just need to be pointed in the right direction. My problem is the following:

$$ \dfrac{\sec(x)-\cos(x)}{\tan(x)} $$

Here's my work trying to solve the problem. Did I make a mistake somewhere or make it more complicated than it needed to be?

$$ \dfrac{\sec(x)}{\tan(x)}-\dfrac{\cos(x)}{\tan(x)}=\dfrac{\dfrac{1}{\cos(x)}}{\dfrac{\sin(x)}{\cos(x)}}-\dfrac{\cos(x)}{\dfrac{\sin(x)}{\cos(x)}}=\dfrac{1}{\cos(x)}*\dfrac{\cos(x)}{\sin(x)}-\dfrac{\cos(x)}{\tan(x)}=\dfrac{1}{\sin(x)}-\dfrac{\cos(x)}{\tan(x)} $$

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Since you want a hint, you're almost there. Now expand $\frac {\cos x}{\tan x }$ amd then combine the two fractions. After that, use a well known trigonometric identity, and you're done!

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  • $\begingroup$ Aw man, I just gave up to soon. I thought for sure I was doing it wrong. Thanks for the tip! $\endgroup$ – Tanner Mar 4 '14 at 3:06
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Multiply the top and bottom by $\cos(x)$. Then life become simpler and happier.

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HINT: Write $\sec(x)$ as $1/\cos(x)$. Try to use $\sin^2(x) = 1 - \cos^2(x)$. Enjoy. Happy mathing.

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Write everything in terms of $\sin x$ and $\cos x$ (which means you should get rid of $\sec x$ and $\tan x$). Gather together fractions over a common denominator. Simplify. Use $\sin^x + \cos^2x = 1$.

You got rid of $\tan x$, but then you brought it back again, which was counter-productive.

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