1. **State the problem:** Prove the trigonometric identity $$\tan \theta + \cot \theta \equiv \frac{2}{\sin 2\theta}$$. 2. **Recall definitions:** - $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ - $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$ - Double angle formula: $$\sin 2\theta = 2 \sin \theta \cos \theta$$ 3. **Start with the left-hand side (LHS):** $$\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$$ 4. **Find common denominator and combine:** $$= \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$$ 5. **Use Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ 6. **Simplify numerator:** $$= \frac{1}{\sin \theta \cos \theta}$$ 7. **Express denominator using double angle formula:** Since $$\sin 2\theta = 2 \sin \theta \cos \theta$$, then $$\sin \theta \cos \theta = \frac{\sin 2\theta}{2}$$ 8. **Substitute back:** $$\frac{1}{\sin \theta \cos \theta} = \frac{1}{\frac{\sin 2\theta}{2}} = \frac{2}{\sin 2\theta}$$ 9. **Conclusion:** $$\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$$, which proves the identity. **Final answer:** $$\boxed{\tan \theta + \cot \theta \equiv \frac{2}{\sin 2\theta}}$$