Question

If $\sin B=3\sin (2A+B),$ then $2\tan A+\tan (A+B)$ =

• A. $0$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$0$

Answer 2

Given that, $\sin B=3\sin (2A+B)$
$\Rightarrow$ $\frac{\sin \,\,B}{\sin \,(2A+B)}=\frac{3}{1}$
$\Rightarrow$ $\frac{\sin B+\sin (2A+B)}{\sin B-\sin (2A+B)}=\frac{3+1}{3-1}$
(use componendo-dividendo formula)
$\Rightarrow$ $-\frac{2\sin \,(A+B).cos(A)}{2\cos \,(A+B).\sin (A)}=\frac{4}{2}$
$\Rightarrow$ $\tan (A+B).\cot \,A=-2$
$\Rightarrow$ $\tan \,(A+B)=-2\tan A$
$\Rightarrow$ $2\tan A+\tan (A+B)=0$