Question

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

• A. $\tan 2A=\tan B$
• B. $\tan 2A={\tan }^{2}B$
• C. $\tan 2A={\tan }^{2}B+2\tan B$
• D. none of these

Answer 1

Answer: (A)

$\tan 2A=\tan B$

$\tan A=\frac{1-\cos B}{\sin B}=\frac{2{\sin }^2\frac{B}{2}}{2\sin \frac{B}{2}\cos \frac{B}{2}}$ ....... $(\because$ $1-cos2x=2si{n}^{2}x)$ and $[\because \sin 2x=2\sin x\cos x]$

$\Rightarrow \tan A=\tan \frac{B}{2}$ ....... $[\frac{\sin \theta }{\cos \theta }=\tan \theta ]$

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

$=\frac{2\tan \frac{B}{2}}{1-{\tan }^2\frac{B}{2}}$
$\Rightarrow \tan 2A=\tan B$
Ans: A