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.

Note that the given expression implies:
$\tan(A)=\frac{1-\cos(B)}{\sin(B)} \cdot \frac{1+\cos(B)}{1+\cos(B)}=\frac{\sin(B)}{1+\cos(B)}$
Therefore
$\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}=\frac{2\frac{\sin(B)}{1+\cos(B)}}{1-\left(\frac{1-\cos(B)}{\sin(B)}\cdot \frac{\sin(B)}{1+\cos(B)}\right)}=\frac{2\sin(B)}{2\cos(B)}=\tan(B)$
