Show that \(\frac{\tan \theta}{1 + \cos \theta} + \frac{\tan \theta}{1 - \cos \theta} \equiv \frac{2}{\sin \theta \cos \theta}\).
Solution
Start with the left-hand side:
\(\frac{\tan \theta}{1 + \cos \theta} + \frac{\tan \theta}{1 - \cos \theta} = \frac{\tan \theta (1 - \cos \theta) + \tan \theta (1 + \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)}\)
\(= \frac{\tan \theta (1 - \cos^2 \theta)}{\sin^2 \theta}\)
\(= \frac{2 \tan \theta}{\sin^2 \theta}\)
\(= \frac{2 \sin \theta}{\cos \theta \sin^2 \theta}\)
\(= \frac{2}{\sin \theta \cos \theta}\)
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