Show that \(\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta} \equiv \frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta}\).
Solution
Start by combining the fractions:
\(\frac{(\tan \theta + 1)(1 - \cos \theta) + (\tan \theta - 1)(1 + \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)}\)
This simplifies to:
\(\frac{\tan \theta - \tan \theta \cos \theta + 1 - \cos \theta + \tan \theta + \tan \theta \cos \theta - 1 - \cos \theta}{1 - \cos^2 \theta}\)
Combine like terms:
\(\frac{2\tan \theta - 2\cos \theta}{1 - \cos^2 \theta}\)
Since \(1 - \cos^2 \theta = \sin^2 \theta\), the expression becomes:
\(\frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta}\)
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