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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.

  • 0
    @HenningMakholm ah right, sorry, I forgot.2012-07-17

3 Answers 3

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$RHS = \frac{\tan A}{1-\tan^2 A} = \frac{\frac{\sin A}{\cos A}}{1-\frac{\sin^2 A}{\cos^2 A}}$ $=\frac{\frac{\sin A}{\cos A}}{\frac{\cos^2 A - \sin^2 A}{\cos^2 A}}\cdot \frac{\cos^2 A}{\cos^2 A}$ $=\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = LHS$

  • 0
    A perfectly explained solution.2012-07-16
2

Your error lies in how you simplified the right hand side (the denominator specifically). Try again! Turn $\tan x$ into $\sin x$ and $\cos x$ with $\displaystyle\tan x =\frac{\sin x}{\cos x}$.

Now, $\frac{\tan{x}}{1-\tan^2{x}}=\frac{\left(\frac{\sin x}{\cos x}\right)}{1-\left(\frac{\sin x}{\cos x}\right)^2}$ You can multiply an expression by 1 and not change the value, (Since $1\cdot a=a$). Now, the problem is which 1 do you multiply by?

You can achieve this by multiplying and distributing by $\displaystyle \frac{\cos^2x}{\cos^2x}$.

  • 0
    And the Pythagorean Identities part can be expanded to include $\cos^2x=1-\sin^2x=(1+\sin x)(1-\sin x)$. Anytime you see things like that try to find a Pythagorean identity2012-07-16
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Use $\sin 2A= 2\sin A \cos A$ and $ \cos 2A= \cos^2 A- \sin^2 A$ to get $ \frac{\sin A \cos A}{\cos^2 A- \sin^2 A}=\frac{\sin 2A}{2\cos 2A}=\frac12\tan{2A}, $ which is equivalent to $ \frac{ \tan A} {1 - \tan^2 A} . $

  • 0
    See [here](http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-.2C_triple-.2C_and_half-angle_formulae) for reference.2012-07-16