1. **Stating the problem:** We want to prove that $$\tan \alpha + \cot \alpha = \frac{1}{\sin \alpha \cos \alpha}$$ for an angle $\alpha$. 2. **Recall definitions:** - $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$ - $\cot \alpha = \frac{\cos \alpha}{\sin \alpha}$ 3. **Express the left side using these definitions:** $$\tan \alpha + \cot \alpha = \frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}$$ 4. **Find a common denominator:** $$= \frac{\sin^2 \alpha}{\sin \alpha \cos \alpha} + \frac{\cos^2 \alpha}{\sin \alpha \cos \alpha} = \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}$$ 5. **Use the Pythagorean identity:** $$\sin^2 \alpha + \cos^2 \alpha = 1$$ 6. **Substitute back:** $$\tan \alpha + \cot \alpha = \frac{1}{\sin \alpha \cos \alpha}$$ **Final answer:** $$\boxed{\tan \alpha + \cot \alpha = \frac{1}{\sin \alpha \cos \alpha}}$$ This proves the given identity using fundamental trigonometric definitions and the Pythagorean identity.