Comment
Share
Q)

# If $\tan A=\large\frac{1-\cos B}{\sin B}$,then $\tan 2A$=_______

$\begin{array}{1 1}(A)\;\large\frac{1}{2}&(B)\;\tan B\\(C)\;1&(D)\;\tan 2B\end{array}$

Comment
A)
Toolbox:
• $\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
Given :
$\tan A=\large\frac{1-\cos B}{\sin B}$
$\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\Rightarrow \large\frac{2\Large\frac{(1-\cos B)}{\sin B}}{1-\Large\frac{(1-\cos B)^2}{\sin ^2B}}$
$\Rightarrow \large\frac{2(1-\cos B)}{\sin B}\times \frac{\sin ^2B}{\sin ^2B-(1-\cos B)^2}$
$\Rightarrow \large\frac{2(1-\cos B)\times \sin B}{\sin ^2B-(1+\cos ^2B-2\cos B)}$
$\Rightarrow \large\frac{2(1-\cos B)\times \sin B}{\sin^2B-1-\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2(1-\cos B)\times \sin B}{\sin ^2B-\sin ^2B-\cos ^2B-\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2(1-\cos B)\times \sin B}{-2\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2(1-\cos B)\times \sin B}{2\cos B(1-\cos B)}$
$\Rightarrow \large\frac{\sin B}{\cos B}$$=\tan B$
Hence (B) is the correct answer.