Prove the identity \(\frac{1 - 2 \sin^2 \theta}{1 - \sin^2 \theta} \equiv 1 - \tan^2 \theta\).
Solution
We start with the left-hand side of the identity:
\(\frac{1 - 2 \sin^2 \theta}{1 - \sin^2 \theta}\).
Recognize that \(1 - \sin^2 \theta = \cos^2 \theta\) using the Pythagorean identity.
Substitute to get:
\(\frac{1 - 2 \sin^2 \theta}{\cos^2 \theta}\).
Rewrite \(1 - 2 \sin^2 \theta\) as \(\cos^2 \theta - \sin^2 \theta\) using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\).
Thus, the expression becomes:
\(\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}\).
Separate the fraction:
\(\frac{\cos^2 \theta}{\cos^2 \theta} - \frac{\sin^2 \theta}{\cos^2 \theta}\).
This simplifies to:
\(1 - \tan^2 \theta\).
Thus, the identity is proven.
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