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Algebra Difficulty 5.0 AIME, harder Find the answer

Compute tan⁡2(20∘)−sin⁡2(20∘)tan⁡2(20∘)sin⁡2(20∘)\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)}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

If we multiply top and bottom by cos⁡2(20∘)\cos ^{2}\left(20^{\circ}\right), the numerator becomes sin⁡2(20∘)⋅(1−cos⁡220∘)=sin⁡4(20∘)\sin ^{2}\left(20^{\circ}\right) \cdot(1-\cos ^{2} 20^{\circ})=\sin ^{4}\left(20^{\circ}\right), while the denominator becomes sin⁡4(20∘)\sin ^{4}\left(20^{\circ}\right) also. So they are equal, and the ratio is 1.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.