Prove the identity \(\frac{\sin x}{1 - \sin x} - \frac{\sin x}{1 + \sin x} \equiv 2 \tan^2 x\).
Solution
Let \(s = \sin x\) and \(c = \cos x\).
Start with the left-hand side:
\(\frac{s}{1 - s} - \frac{s}{1 + s}\)
Combine the fractions:
\(\frac{s(1 + s) - s(1 - s)}{(1 - s)(1 + s)} = \frac{s + s^2 - s + s^2}{1 - s^2} = \frac{2s^2}{1 - s^2}\)
Use the identity \(1 - s^2 = c^2\):
\(\frac{2s^2}{c^2}\)
Since \(\tan x = \frac{s}{c}\), we have:
\(\frac{2s^2}{c^2} = 2 \left( \frac{s}{c} \right)^2 = 2 \tan^2 x\)
Thus, the identity is proven.
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