Question

If $\frac{\sin 2(A+B)+\sin 2B}{\sin 2A}=2,$ then $\tan (A+B)$ is equal to

• A. $3\tan A$
• B. $3\tan B$
• C. $\tan A$
• D. $\tan B$

Answer 1

The correct option is B $3\tan B$

$\frac{\sin 2(A+B)+\sin 2B}{\sin 2A}=2\Rightarrow \frac{2\sin (A+2B)\cos A}{2\sin A\cos A}=2$

$\Rightarrow \frac{\sin (A+2B)}{\sin A}=2$

Hence $\frac{\sin (A+2B)+\sin A}{\sin (A+2B)-\sin A}=3$

$\Rightarrow \frac{2\sin (A+B)\cos B}{2\cos (A+B)\sin B}=3\Rightarrow \tan$ $(A+B)=3\tan B$