<s>[INST] 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.
 [/INST] Continuing from your last step,
    $$
    LHS=\frac{\frac{2-2\cos B}{\sin B}}{1-\frac{(1-\cos B)^2}{\sin^2B}}\\
       =\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2B-(1-\cos B)^2}{\sin^2B}} \\
       =\frac{2-2\cos B}{\frac{\sin^2B-1-\cos^2B+2\cos B}{\sin B}}
      $$
and then use that $1=\sin^2+\cos^2B$ to get
  $$
 LHS= \frac{2(1-\cos B)(\sin B)}{\sin^2B-\sin^2B-\cos^2B-\cos^2B+2\cos B}\\
    =\frac{2(1-\cos B)(\sin B)}{2\cos B(1-\cos B)}\\
    =\frac{\sin B}{\cos B}=\tan B=RHS
$$
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