Trig Reduction with Pythagoras If $\sin 10 = p$, then determine $$\tan^2 30^\circ \times \tan^2190^\circ$$ in terms of p.

$\tan^2 30^\circ$ is easy because it's a common angle; we know that it is equal to $\frac{1}{3}$
Note that $\tan^2 190^\circ = \tan^2 10^\circ = \frac{\sin^2 10^\circ}{\cos^2 10^\circ}$
Since $\cos^2 \theta = 1 - \sin^2 \theta$, we find that $\tan^2 190^\circ = \frac{p^2}{1 - p^2}$
Thus, the final answer is $\frac{p^2}{3(1 - p^2)}$
