If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$ Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt] &=\frac{2\tan A}{1-\tan ^2A} \\[6pt] &=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1-\cos B)^2}{\sin^2 B}} \end{align}$$
On simplification from here, I could not get the required R.H.S.