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prove trig identity...
\[\tan \theta=\frac{ \sin 2\theta }{ 1+\cos2\theta }\]
write sin 2theta = 2sin theta cos theta and 1+cos 2 theta = 2 cos^2 theta.
We already know that,\[\tan = {\sin \over \cos}\]and if you try to simplify\[2\sin\theta \cos\theta \over 2\cos^2\theta\], then you get \(\large{\sin\theta\over \cos\theta}\)