If \(\sin A=3 \sin (A+2 B)\) angle \(B\) is acute and \(A\) is obtuse: then (a) \(\tan \mathrm{B}=(1 / \sqrt{2})\) (b) \(\tan B>(1 / \sqrt{2})\) (c) \(\tan \mathrm{B}<(1 / \sqrt{2})\) (d) \(0<\tan B<(1 / \sqrt{2})\)

Use the identity \(\sin(A+2B) = \sin A \cos 2B + \cos A \sin 2B\) to rewrite the given equation: \(\sin A = 3[\sin A \cos 2B + \cos A \sin 2B]\)

Use the double angle identities \(\sin(2B) = 2\sin B \cos B\) and \(\cos(2B) = \cos^2 B - \sin^2 B\): \(\sin A = 3[\sin A (\cos^2 B - \sin^2 B) + \cos A (2\sin B \cos B)]\)

Since we know that \(A\) is obtuse, \(\sin A \neq 0\). Thus, we can divide both sides of the equation by \(\sin A\): \(1 = 3[\cos^2 B - \sin^2 B + \frac{\cos A}{\sin A}(2\sin B \cos B)]\)

Now we introduce \(\tan B = \frac{\sin B}{\cos B}\): \(1 = 3[(1 - \tan^2 B) - \tan^2 B + \frac{2\tan B}{\sin A \cos A} (1 - \tan^2 B)]\)

Rearrange the equation to isolate \(\tan B\): \(\tan^2 B (3 - 3 \frac{2}{\sin A\cos A}) + \tan B ( 6\frac{1}{\sin A\cos A}) - 3 = 0\) Since the equation is quadratic in \(\tan^2 B\) and \(B\) is acute, we need to find \(\tan^2 B\) and take the square root. Unfortunately, solving the quadratic equation above directly does not yield an easy solution for \(\tan B\).

At this point, we notice that our task is not to solve for \(\tan B\), but rather to compare the obtained expression for \(\tan B\) with the given options. Thus, we can analyze each option to see if it leads to a consistent equation for our derived quadratic equation. (a) If \(\tan B = \frac{1}{\sqrt{2}}\), then \(\tan^2 B = \frac{1}{2}\), which leads to an inconsistent equation in the above quadratic equation. (b) If \(\tan B > \frac{1}{\sqrt{2}}\), then \(\tan B < \sqrt{2}\), it does not provide sufficient information to uniquely solve our quadratic equation. (c) If \(\tan B < \frac{1}{\sqrt{2}}\), then \(\tan B > \sqrt{2}\), and we again cannot uniquely solve our quadratic equation. (d) If \(0 < \tan B < \frac{1}{\sqrt{2}}\), this is consistent with what we have derived so far in our equation. We can conclude that the correct option is: (d) \(0 < \tan B < \frac{1}{\sqrt{2}}\)