### Instruction: In right triangle $DEF$, we have $\angle E = 90^\circ$ and $\tan D = 3\sin D$.  What is $\sin F$?### Response: I know that $\tan D = \frac{\sin D}{\cos D}$, so I can rewrite the given equation as $\frac{\sin D}{\cos D} = 3\sin D$.
Multiplying both sides by $\cos D$, I get $\sin D = 3\sin D \cos D$.
Dividing both sides by $\sin D$, I get $1 = 3\cos D$.
Solving for $\cos D$, I get $\cos D = \frac{1}{3}$.
Now I can use the Pythagorean identity $\sin^2 D + \cos^2 D = 1$ to find $\sin D$.
Substituting $\cos D = \frac{1}{3}$, I get $\sin^2 D + \frac{1}{9} = 1$.
Subtracting $\frac{1}{9}$ from both sides, I get $\sin^2 D = \frac{8}{9}$.
Taking the positive square root, I get $\sin D = \frac{2\sqrt{2}}{3}$.
To find $\sin F$, I can use the fact that the sum of the angles in a triangle is $180^\circ$.
So $\angle F = 180^\circ - \angle D - \angle E = 180^\circ - \angle D - 90^\circ = 90^\circ - \angle D$.
Therefore, $\sin F = \sin(90^\circ - \angle D) = \cos D$.
But I already know that $\cos D = \frac{1}{3}$, so $\sin F = \frac{1}{3}$.