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.
 A: 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.
A: 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
$\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}=\frac{2\frac{\sin(B)}{1+\cos(B)}}{1-\left(\frac{1-\cos(B)}{\sin(B)}\cdot \frac{\sin(B)}{1+\cos(B)}\right)}=\frac{2\sin(B)}{2\cos(B)}=\tan(B)$
A: 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}
A: 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
  $$
 LHS= \frac{2(1-\cos B)(\sin B)}{\sin^2B-\sin^2B-\cos^2B-\cos^2B+2\cos B}\\
    =\frac{2(1-\cos B)(\sin B)}{2\cos B(1-\cos B)}\\
    =\frac{\sin B}{\cos B}=\tan B=RHS
$$
