If $ \sin B=3 \sin (2A+B)$, prove that $2\tan A+\tan (A+B)=0$ 
Given $\sin B=3\sin(2A+B)$, prove $ 2\tan A+\tan(A+B)=0$.

My book uses componendo and dividendo approach to do this which I feel is bit unintuitive. I tried to do this by using identity for $\sin(x+y)=\sin x\cos y+\cos x\sin y$ but could not reach to answer. How do I do this?

The hypothesis can be written :
$$\sin((A+B)-A)=3\sin((A+B)+A)$$
Now using the formula you mention :
$$\sin(A+B)\cos(A)-\cos(A+B)\sin(A)=3\left(\sin(A+B)\cos(A)+\cos(A+B)\sin(A)\right)$$
which simplifies to :
$$\sin(A+B)\cos(A)+2\cos(A+B)\sin(A)$$
Finally (assuming $\cos(A)\neq0$ and $\cos(A+B)\neq0$) :
$$\tan(A+B)+2\tan(A)=0$$
