Question

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

• A. $0$
• B. $-1$
• C. $2$
• D. $1$

Answer 1

The correct option is A $0$

$\sin B=3\sin (2A+B)$

$\frac{\sin (2A+B)}{\sin B}=\frac{1}{3}$

$\frac{\sin \left[\right(A+B)+A]}{\sin \left[\right(A+B)-A]}=\frac{1}{3}$

$\frac{\sin \left[\right(A+B)+A]+\sin \left[\right(A+B)-A]}{\sin \left[\right(A+B)+A]-\sin \left[\right(A+B)-A]}=\frac{1+3}{1-3}$

$\frac{\left[\sin \right(A+B)\cos A+\cos (A+B\left)\sin A\right]+\left[\sin \right(A+B)\cos A-\cos (A+B\left)\sin A\right]}{\left[\sin \right(A+B)\cos A+\cos (A+B\left)\sin A\right]-\left[\sin \right(A+B)\cos A-\cos (A+B\left)\sin A\right]}=\frac{1+3}{1-3}$

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

$\tan (A+B)\cot A=-2$

$\tan (A+B)=-2\tan A$

$2\tan A+\tan (A+B)=0$