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

Given $\sin A = n \sin (A+2B)$, rewrite the angles using compound angle formulas.

Expressing in terms of $A+B$ and $B$:
$\sin A = \sin((A+B)-B) = \sin(A+B)\cos B - \cos(A+B)\sin B$
$\sin(A+2B) = \sin((A+B)+B) = \sin(A+B)\cos B + \cos(A+B)\sin B$

Substitute into the original equation:
$\sin(A+B)\cos B - \cos(A+B)\sin B = n(\sin(A+B)\cos B + \cos(A+B)\sin B)$

Rearrange terms to isolate factors:
$\sin(A+B)\cos B(1 - n) = \cos(A+B)\sin B(1 + n)$

Divide both sides by $\cos(A+B)\cos B$:
$\tan(A+B)(1 - n) = \tan B(1 + n)$

Solving for $\tan(A+B)$ yields:
$\tan(A+B) = \frac{1+n}{1-n} \tan B$

Final answer: $\boxed{\text{D}}$