### Instruction: Simplify
\[\frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x}.\].### Response: This problem asks me to simplify a trigonometric expression involving fractions with different denominators.
One possible strategy is to find a common denominator for the fractions and combine them into one term.
To do this, I can multiply the first fraction by \(\frac{1 + \sin x}{1 + \sin x}\) and the second fraction by \(\frac{1 - \sin x}{1 - \sin x}\).
This way, both fractions will have the same denominator of \((1 - \sin x)(1 + \sin x)\), which is also equal to \(1 - \sin^2 x\) by the difference of squares formula.
The result of multiplying the fractions is

\[\frac{\cos x (1 + \sin x)}{1 - \sin^2 x} - \frac{\cos x (1 - \sin x)}{1 - \sin^2 x}.\]
Now I can combine the numerators by subtracting the terms.
This gives

\[\frac{\cos x (1 + \sin x) - \cos x (1 - \sin x)}{1 - \sin^2 x}.\]
Expanding the brackets and simplifying, I get

\[\frac{\cos x + \cos x \sin x - \cos x + \cos x \sin x}{1 - \sin^2 x}.\]
The \(\cos x\) terms cancel out, leaving

\[\frac{2 \cos x \sin x}{1 - \sin^2 x}.\]
This looks simpler than the original expression, but I wonder if I can simplify it further.
I recall that there is a double angle identity for \(\sin 2x\), which is \(\sin 2x = 2 \sin x \cos x\).
This means that I can rewrite the numerator as \(\sin 2x\).
The denominator is also a trigonometric identity, since \(1 - \sin^2 x = \cos^2 x\).
Therefore, the final simplified expression is

\[\frac{\sin 2x}{\cos^2 x}.\]