An interesting problem about circle

With the angles as labelled, note that $\angle CBD = 90^\circ-\theta$ and $\angle CBE = 90^\circ - \phi$ . A couple of applications of the Sine Rule tell us that $\frac{\sin\theta}{3} = \frac{\cos\phi}{CE}$ and $\frac{\sin\phi}{3} = \frac{\cos\theta}{CD}$ , so that $\frac{\sin\theta \sin\phi}{9} \; = \; \frac{\cos\theta \cos\phi}{CE \times CD}$

The Intersecting Chords Theorem tells us that $CD \times CE \; = \; CB \times CF \; = \; 3 \times 7 \; = \; 21$ and hence $\tan\theta \tan\phi \; = \; \frac{9}{CD \times CE} \; = \; \boxed{\frac{3}{7}}$