Question

If $\tan A=\frac{1-\cos B}{\sin B}$, then the value of $\frac{2\tan A}{1-{\tan }^{2}A}$ is

• A. $\frac{\left(\tan B\right)}{2}$
• B. $2\tan B$
• C. $\tan B$
• D. $4\tan B$

Answer 1

The correct option is C $\tan B$

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

$=\frac{2{\sin }^2\frac{B}{2}}{2\sin \frac{B}{2}\cdot \cos \frac{B}{2}}$

$=\tan \frac{B}{2}$

Therefore, $A=\frac{B}{2}\Rightarrow 2A=B$

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