Problem

16 Prove that $\tan \theta+\frac{1}{\tan \theta} \equiv \frac{1}{\sin \theta \cos \theta}$

Answer

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Answer

\(\boxed{\tan \theta + \frac{1}{\tan \theta} \equiv \frac{1}{\sin \theta \cos \theta}}\)

Steps

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 ：\(\boxed{\tan \theta + \frac{1}{\tan \theta} \equiv \frac{1}{\sin \theta \cos \theta}}\)

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