Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$ 
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.

You can proceed in this way:
$$\tan A=\frac{1-\cos B}{\sin B}=\frac{2\sin^2 \frac{B}{2}}{2\sin\frac{B}{2}\cos\frac{B}{2}}=\frac{\sin\frac{B}{2}}{\cos\frac{B}{2}}=\tan \frac{B}{2}$$
And hence comparing, we can write that $A=n\pi + \frac{B}{2}$ where $n$ is any integer.
So we can say that $2A=2n\pi + B \Rightarrow \tan 2A = \tan(2n\pi + B) = \tan B$
Hence proved.
