Question

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

Answer 1

Simplify the given expression by using multiple angles formula

Given:$\tan A=\frac{1-\cos B}{\sin B}$

$\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$

$=\frac{2\left(\frac{1-\cos B}{\sin B}\right)}{1-{\left(\frac{1-\cos B}{\sin B}\right)}^2}$

$\tan 2A=\frac{2\left(\frac{2{\sin }^2\left(\frac{B}{2}\right)}{2\sin \left(\frac{B}{2}\right)\cos \left(\frac{B}{2}\right)}\right)}{1-{\left(\frac{2{\sin }^2\left(\frac{B}{2}\right)}{2\sin \left(\frac{B}{2}\right)\cos \left(\frac{B}{2}\right)}\right)}^2}$

$\left[\begin{array}{c}\because 1-\cos 2\theta =2{\sin }^2\theta \&\\ \sin 2\theta =2\sin \theta \cos \theta \end{array}\right]$

$=\frac{2\tan \left(\frac{B}{2}\right)}{1-{\tan }^2\left(\frac{B}{2}\right)}$

$=\tan B$

$\therefore \tan 2A=\tan B$