1. **Problem statement:** Prove that $$\tan \theta + \cot \theta \equiv 2 \csc 2\theta$$ for $$\theta \neq \frac{n\pi}{2}, n \in \mathbb{Z}$$. 2. **Recall definitions and identities:** - $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ - $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$ - $$\csc 2\theta = \frac{1}{\sin 2\theta}$$ - 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:** $$= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$$ 5. **Use Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ 6. **So LHS becomes:** $$\frac{1}{\sin \theta \cos \theta}$$ 7. **Express denominator using double angle formula:** $$\sin 2\theta = 2 \sin \theta \cos \theta \implies \sin \theta \cos \theta = \frac{\sin 2\theta}{2}$$ 8. **Substitute back:** $$\frac{1}{\sin \theta \cos \theta} = \frac{1}{\frac{\sin 2\theta}{2}} = \frac{2}{\sin 2\theta} = 2 \csc 2\theta$$ 9. **Therefore:** $$\tan \theta + \cot \theta \equiv 2 \csc 2\theta$$ --- 10. **Explain why $$\tan \theta + \cot \theta = 1$$ has no real solutions:** From the proven identity, $$\tan \theta + \cot \theta = 2 \csc 2\theta$$. So the equation becomes: $$2 \csc 2\theta = 1 \implies \csc 2\theta = \frac{1}{2}$$ Since $$\csc x = \frac{1}{\sin x}$$, this means: $$\frac{1}{\sin 2\theta} = \frac{1}{2} \implies \sin 2\theta = 2$$ But the sine function only takes values in $$[-1,1]$$, so $$\sin 2\theta = 2$$ is impossible. Hence, the equation $$\tan \theta + \cot \theta = 1$$ has no real solutions.