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How would I simplify this difficult trigonometric identity:

$$\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}.$$

I am not exactly sure what to do.

I simplified the right side to

$$\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos^2 A}{\sin^2 A}}$$

But how would I proceed.

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    Use $\sin 2A= 2\sin A \cos A$ and $ \cos 2A= \cos^2 A- \sin^2 A$.2012-07-16
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    I will say all the post were helpful and I would give everyone recognition were I a registered user.2012-07-16
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    James, you can up/down vote on all posts (question, answers, comments) here (except your own ;-). Additionally you can choose among the given answer and accept the one that suits you most. Read the [faq](http://math.stackexchange.com/faq) for more information. And last: Welcome to Math.StackExchange.com...2012-07-16
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    @draks, the ability to upvote requires 15 rep, which James had had for only about 2 minutes when he posted that comment. Quite possibly he hadn't noticed yet. _Downvoting_, on the other hand, requires 125 rep.2012-07-16
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    @HenningMakholm ah right, sorry, I forgot.2012-07-17

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