9709 P12 - Mar 2023 - Q7B - 3 marks
502
Show that \(\frac{\tan \theta}{\sin \theta} - \frac{\sin \theta}{\tan \theta} \equiv \tan \theta \sin \theta\).
Solution
Checked by expert
Start with the left-hand side:
\(\frac{\tan \theta}{\sin \theta} - \frac{\sin \theta}{\tan \theta}\)
Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\(\frac{\frac{\sin \theta}{\cos \theta}}{\sin \theta} - \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}}\)
Simplify each term:
\(\frac{1}{\cos \theta} - \sin \theta \cos \theta\)
Combine the terms over a common denominator:
\(\frac{1 - \cos^2 \theta}{\cos \theta}\)
Use the identity \(1 - \cos^2 \theta = \sin^2 \theta\):
\(\frac{\sin^2 \theta}{\cos \theta}\)
Recognize this as \(\tan \theta \sin \theta\):
\(\tan \theta \sin \theta\)
