Prove that \( \tan \theta+\cot \theta=\frac{1}{\cos \theta \sin \theta} \)

To prove the identity \( \tan \theta+\cot \theta=\frac{1}{\cos \theta \sin \theta} \), we start with the left side of the equation: \[ \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \] To combine these fractions, we find a common denominator, which is \( \sin \theta \cos \theta \): \[ = \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta} \] Now we combine the fractions: \[ = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \] Using the Pythagorean identity, we know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Therefore, we have: \[ = \frac{1}{\sin \theta \cos \theta} \] The right side of the equation was \( \frac{1}{\cos \theta \sin \theta} \). Thus: \[ \tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta} = \frac{1}{\cos \theta \sin \theta} \] This completes the proof, confirming that: \[ \tan \theta + \cot \theta = \frac{1}{\cos \theta \sin \theta} \] Hence, the identity is proven.