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

To establish the identity: \[ \frac{\tan \theta + \cot \theta}{\tan \theta - \cot \theta} = \frac{1}{\sin^{2} \theta - \cos^{2} \theta}, \] we'll start by expressing \(\tan \theta\) and \(\cot \theta\) in terms of sine and cosine. 1. **Express \(\tan \theta\) and \(\cot \theta\):** \[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}. \] 2. **Compute the Numerator \(\tan \theta + \cot \theta\):** \[ \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}, \] since \(\sin^2 \theta + \cos^2 \theta = 1\). 3. **Compute the Denominator \(\tan \theta - \cot \theta\):** \[ \tan \theta - \cot \theta = \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}. \] 4. **Form the Ratio:** \[ \frac{\tan \theta + \cot \theta}{\tan \theta - \cot \theta} = \frac{\frac{1}{\sin \theta \cos \theta}}{\frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}} = \frac{1}{\sin^2 \theta - \cos^2 \theta}. \] Thus, the identity is verified: \[ \frac{\tan \theta + \cot \theta}{\tan \theta - \cot \theta} = \frac{1}{\sin^{2} \theta - \cos^{2} \theta}. \] **Answer:** By rewriting tan and cot in terms of sine and cosine and simplifying, both sides reduce to 1 ⁄ (sin²θ – cos²θ). Thus, the identity is established.