If, $\tan A=\dfrac{1-cos~B}{\sin B},$ then what is $\dfrac{2 \tan A}{1-{{\tan }^{2}}A}$ equal to ?
Solution
$= \dfrac{2\left(\dfrac{1 - \cos B}{\sin B}\right)}{1 - \left(\dfrac{1 - \tan \cos B}{\sin B}\right)^2} = \dfrac{2 \left(\dfrac{1 - \cos B}{\sin B}. \sin B\right)}{\sin^2 B [1 + \cos^2 B - 2 \cos B]}$
$= \dfrac{2 (1 - \cos B).\sin B}{\sin^2 B - 1 - \cos^2 B + 2 \cos B} = \dfrac{2 (1 - \cos B)sin B}{\sin^2 B - \sin^2 B - \cos^2 B - \cos^2 B + 2 \cos B}$
$= \dfrac{2((1 - \cos B).\sin B}{-2 \cos^2 B + 2 \cos B} = tan B$