Prove the identity. \[ \frac{\cos ^{2} \alpha+\cot \alpha}{\cos ^{2} \alpha-\cot \alpha}=\frac{\cos ^{2} \alpha \tan \alpha+1}{\cos ^{2} \alpha \tan \alpha-1} \] To verify the identity, work each side separately until you obtain the same expression. Start with the left side. Choose the correct step and transform the expression according to the step chosen. \( \frac{\cos ^{2} \alpha+\cot \alpha}{\cos ^{2} \alpha-\cot \alpha}=\square \)

To verify the identity, we start with the left side: \[ \frac{\cos^{2} \alpha + \cot \alpha}{\cos^{2} \alpha - \cot \alpha} \] We know that: \[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \] Substituting this into the expression gives us: \[ = \frac{\cos^{2} \alpha + \frac{\cos \alpha}{\sin \alpha}}{\cos^{2} \alpha - \frac{\cos \alpha}{\sin \alpha}} \] Now, to combine the fractions, we will multiply both the numerator and denominator by \(\sin \alpha\): \[ = \frac{\sin \alpha (\cos^{2} \alpha) + \cos \alpha}{\sin \alpha (\cos^{2} \alpha) - \cos \alpha} \] This transforms our expression to: \[ = \frac{\sin \alpha \cos^{2} \alpha + \cos \alpha}{\sin \alpha \cos^{2} \alpha - \cos \alpha} \] Next, factor out a \(\cos \alpha\) in the numerator and the denominator: \[ = \frac{\cos \alpha \left( \sin \alpha \cos \alpha + 1 \right)}{\cos \alpha \left( \sin \alpha \cos \alpha - 1 \right)} \] Cancel the \(\cos \alpha\) (assuming \(\cos \alpha \neq 0\)): \[ = \frac{\sin \alpha \cos \alpha + 1}{\sin \alpha \cos \alpha - 1} \] Now we have simplified the left side to: \[ \frac{\sin \alpha \cos \alpha + 1}{\sin \alpha \cos \alpha - 1} \] Next, we should simplify the right side starting from: \[ \frac{\cos^{2} \alpha \tan \alpha + 1}{\cos^{2} \alpha \tan \alpha - 1} \] Using \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\), we substitute: \[ = \frac{\cos^{2} \alpha \cdot \frac{\sin \alpha}{\cos \alpha} + 1}{\cos^{2} \alpha \cdot \frac{\sin \alpha}{\cos \alpha} - 1} \] This simplifies to: \[ = \frac{\sin \alpha \cos \alpha + 1}{\sin \alpha \cos \alpha - 1} \] As we can see, both sides now reduce to: \[ \frac{\sin \alpha \cos \alpha + 1}{\sin \alpha \cos \alpha - 1} \] Thus, we have verified the identity. Therefore, the identity is proven to be true.