Establish the following identity. \[ \frac{\tan \theta+\cot \theta}{\tan \theta-\cot \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta} \]

To prove the identity \[ \frac{\tan \theta+\cot \theta}{\tan \theta-\cot \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}, \] we start by rewriting the left-hand side: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \cot \theta = \frac{\cos \theta}{\sin \theta}. \] Substituting these into the left-hand side, we get: \[ \frac{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}}{\frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta}}. \] We can find a common denominator for both the numerator and the denominator: The numerator becomes: \[ \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} \quad \text{(since } \sin^2 \theta + \cos^2 \theta = 1\text{)}. \] The denominator becomes: \[ \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}. \] Now the full left-hand side can be simplified: \[ \frac{\frac{1}{\sin \theta \cos \theta}}{\frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}} = \frac{1}{\sin^2 \theta - \cos^2 \theta}. \] We see that both the left-hand side and the right-hand side are indeed the same: \[ \frac{\tan \theta+\cot \theta}{\tan \theta-\cot \theta} = \frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}. \] Thus, the identity is established: \[ \frac{\tan \theta+\cot \theta}{\tan \theta-\cot \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}. \] The identity is verified.