Compute
\[\frac{\tan^2 20^\circ - \sin^2 20^\circ}{\tan^2 20^\circ \sin^2 20^\circ}.\]
We have that
\begin{align*}
\frac{\tan^2 20^\circ - \sin^2 20^\circ}{\tan^2 20^\circ \sin^2 20^\circ} &= \frac{\frac{\sin^2 20^\circ}{\cos^2 20^\circ} - \sin^2 20^\circ}{\frac{\sin^2 20^\circ}{\cos^2 20^\circ} \cdot \sin^2 20^\circ} \\
&= \frac{\sin^2 20^\circ - \cos^2 20^\circ \sin^2 20^\circ}{\sin^4 20^\circ} \\
&= \frac{1 - \cos^2 20^\circ}{\sin^2 20^\circ} = \boxed{1}.
\end{align*}
1
Level 1
Precalculus
Problem: Compute
\[\frac{\tan^2 20^\circ - \sin^2 20^\circ}{\tan^2 20^\circ \sin^2 20^\circ}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we start by rewriting the expression using trigonometric identities. Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Therefore, we can express \(\tan^2 20^\circ\) as:

\[\tan^2 20^\circ = \frac{\sin^2 20^\circ}{\cos^2 20^\circ}.\]

Substituting this into the original expression gives:

\[\frac{\frac{\sin^2 20^\circ}{\cos^2 20^\circ} - \sin^2 20^\circ}{\frac{\sin^2 20^\circ}{\cos^2 20^\circ} \sin^2 20^\circ}.\]

Next, we simplify the numerator:

\[\frac{\sin^2 20^\circ}{\cos^2 20^\circ} - \sin^2 20^\circ = \frac{\sin^2 20^\circ - \sin^2 20^\circ \cos^2 20^\circ}{\cos^2 20^\circ}.\]

Factoring out \(\sin^2 20^\circ\) from the numerator yields:

\[\frac{\sin^2 20^\circ (1 - \cos^2 20^\circ)}{\cos^2 20^\circ} = \frac{\sin^2 20^\circ \sin^2 20^\circ}{\cos^2 20^\circ}.\]

Now, substituting this back into the expression, we have:

\[\frac{\frac{\sin^4 20^\circ}{\cos^2 20^\circ}}{\frac{\sin^4 20^\circ}{\cos^2 20^\circ}}.\]

This simplifies to:

\[\frac{\sin^4 20^\circ}{\cos^2 20^\circ} \cdot \frac{\cos^2 20^\circ}{\sin^4 20^\circ} = 1.\]

Thus, the final answer is \(1\).<|END_OF_LATENT|> The final answer is $1$.