2. Compute $\frac{\tan ^{2}\left(20^{\circ}\right)-\sin ^{2}\left(20^{\circ}\right)}{\tan ^{2}\left(20^{\circ}\right) \sin ^{2}\left(20^{\circ}\right)}$.
Solution: $\square$
If we multiply top and bottom by $\cos ^{2}\left(20^{\circ}\right)$, the numerator becomes $\sin ^{2}\left(20^{\circ}\right) \cdot(1-$ $\left.\cos ^{2} 20^{\circ}\right)=\sin ^{4}\left(20^{\circ}\right)$, while the denominator becomes $\sin ^{4}\left(20^{\circ}\right)$ also. So they are equal, and the ratio is 1 .