Establish the identity. \[ \cos \theta(\cot \theta+\tan \theta)=\boldsymbol{\operatorname { c s c }} \theta \] Write the left side in terms of sine and cosine. \( \cos \theta(\square) \)

To establish the identity \[ \cos \theta(\cot \theta+\tan \theta)=\operatorname{csc} \theta, \] we will start by rewriting the left side in terms of sine and cosine. 1. **Rewrite \(\cot \theta\) and \(\tan \theta\)**: - \(\cot \theta = \frac{\cos \theta}{\sin \theta}\) - \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) 2. **Substituting these into the left side**: \[ \cos \theta \left( \cot \theta + \tan \theta \right) = \cos \theta \left( \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right) \] 3. **Combine the terms inside the parentheses**: To combine the fractions, we need a common denominator, which is \(\sin \theta \cos \theta\): \[ \frac{\cos^2 \theta}{\sin \theta} + \frac{\sin^2 \theta}{\cos \theta} = \frac{\cos^2 \theta \cdot \cos \theta + \sin^2 \theta \cdot \sin \theta}{\sin \theta \cos \theta} = \frac{\cos^3 \theta + \sin^2 \theta \cos \theta}{\sin \theta \cos \theta} \] 4. **Now multiply by \(\cos \theta\)**: \[ \cos \theta \left( \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \right) = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta} \] 5. **Using the Pythagorean identity**: We know that \(\cos^2 \theta + \sin^2 \theta = 1\): \[ \frac{1}{\sin \theta} = \operatorname{csc} \theta \] Thus, we have shown that: \[ \cos \theta(\cot \theta+\tan \theta) = \operatorname{csc} \theta. \] The left side in terms of sine and cosine is: \[ \cos \theta \left( \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right). \]