Problem
16 Prove that $\tan \theta+\frac{1}{\tan \theta} \equiv \frac{1}{\sin \theta \cos \theta}$
Solution
Step 1 ：\(\frac{1}{\sin \theta \cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\)
Step 2 ：\(= \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta}\)
Step 3 ：\(= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\)
Step 4 ：\(= \tan \theta + \frac{1}{\tan \theta}\)
Step 5 ：\(\boxed{\tan \theta + \frac{1}{\tan \theta} \equiv \frac{1}{\sin \theta \cos \theta}}\)
