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I'm must be doing something wrong with this problem, I have used the tanx identity but not getting the right answer. The problem is as follows: verify the following identity:

$$\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$$

Thank you

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    $\begingroup$ I would break it all up into sines and cosines, and use double angle identities on those. You should be able to verify from there that the RHS and LHS are always equal. $\endgroup$ – Arthur Mar 14 '13 at 16:40
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Use $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$

Put $A=B=x$


Alternatively,

as $\sin(A+B)=\sin A\cos B+\cos A\sin B,\sin2x=2\sin x\cos x$(Putting $A+B=x$)

$\cos(A+B)=\cos A\cos B-\sin A\sin B,\cos2x=\cos^2x-\sin^2x$(Putting $A+B=x$)

$$\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\sin x\cos x}{\cos^2x-\sin^2x}$$ $$\tan2x=\frac{2\tan x}{1-\tan^2x}\text{ diving the numerator & denominator by }\cos^2x$$

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