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 are on the right road.... Continuing calculation,
\begin{align}
\frac{\frac{2-2\cos B}{\sin B}}{1-\frac{(1-\cos B)^2}{\sin^2 B}}&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2 B -1+2\cos B-\cos^2 B }{\sin^2 B}}\\
&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2 B -\sin^2 B -\cos^2 B+2\cos B-\cos^2 B }{\sin^2 B}}\\
&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{2\cos B(1-\cos B)}{\sin^2 B}}\\
&=\frac{2\sin^2 B(1-\cos B)}{2\cos B\sin B(1-\cos B)}\\
&=\frac{\sin B}{\cos B}\\
&=\tan B
\end{align}
