Prove the identity \(\frac{\sin \theta}{1 - \cos \theta} - \frac{1}{\sin \theta} \equiv \frac{1}{\tan \theta}\).
Solution
Start with the left-hand side (LHS):
\(\frac{\sin \theta}{1 - \cos \theta} - \frac{1}{\sin \theta}\)
Put over a common denominator:
\(\frac{\sin^2 \theta - (1 - \cos \theta)}{(1 - \cos \theta) \sin \theta}\)
Simplify the numerator using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(\frac{1 - \cos^2 \theta - 1 + \cos \theta}{(1 - \cos \theta) \sin \theta}\)
\(\frac{\cos \theta (1 - \cos \theta)}{(1 - \cos \theta) \sin \theta}\)
Cancel \(1 - \cos \theta\) from the numerator and denominator:
\(\frac{\cos \theta}{\sin \theta}\)
Recognize this as \(\frac{1}{\tan \theta}\).
Log in to record attempts.