9709 P13 - Jun 2012 - Q1I - 3 marks
501
Prove the identity \(\tan^2 \theta - \sin^2 \theta \equiv \tan^2 \theta \sin^2 \theta\).
Solution
Start with the left-hand side: \(\tan^2 \theta - \sin^2 \theta\).
Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).
Substitute into the expression: \(\frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta\).
Factor out \(\sin^2 \theta\): \(\sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right)\).
Using \(1 - \cos^2 \theta = \sin^2 \theta\), rewrite as \(\sin^2 \theta \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)\).
This simplifies to \(\tan^2 \theta \sin^2 \theta\), which is the right-hand side.
