Question

Given $\sin B=\frac{1}{5}\sin (2A+B)$ then, $\tan (A+B)=k\tan A$, where $k$ has the value equal to

• A. $1$
• B. $2$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

$\frac{3}{2}$

$\frac{\sin (2A+B)}{\sin B}=5$
$\frac{\sin (2A+B)+\sin B}{\sin (2A+B)-\sin B}=\frac{5+1}{5-1}$
$\frac{2\sin (A+B)\cos \left(A\right)}{2\sin \left(A\right)\cos (A+B)}=\frac{6}{4}$

$\tan (A+B)=\frac{3}{2}\tan A$
Hence
$k=\frac{3}{2}$