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, we start with the left side: \[ \cos \theta(\cot \theta + \tan \theta). \] Now, recalling that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can rewrite the expression inside the parentheses: \[ \cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}. \] Now finding a common denominator, we get: \[ \cot \theta + \tan \theta = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}, \] because \( \cos^2 \theta + \sin^2 \theta = 1 \). We can now substitute this back into the original expression: \[ \cos \theta \left( \frac{1}{\sin \theta \cos \theta} \right). \] The \( \cos \theta \) in the numerator and denominator cancels out, leading us to: \[ \frac{1}{\sin \theta} = \csc \theta. \] Thus, we have proven that: \[ \cos \theta(\cot \theta + \tan \theta) = \csc \theta. \]