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Today I have to prove a trig identity.

It involves double angle. $$\tan 2\theta + \sin 2\theta = \frac{2\sin 2\theta}{1 - \tan^2\theta}$$

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    $\begingroup$ So... where are you stuck? $\endgroup$ – zahbaz Oct 21 '15 at 5:46
  • $\begingroup$ The Whole Problem Stuck o me $\endgroup$ – Math Boy Oct 21 '15 at 5:47
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$$\tan2y+\sin2y=\sin2y\cdot\dfrac{1+\cos2y}{\cos2y}$$

Now use $\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}$

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