1. **State the problem:** (i) Prove the identity $$\frac{\cos\theta}{\tan\theta(1-\sin\theta)} \equiv 1 + \sin\theta.$$ (ii) Hence solve the equation $$\frac{\cos\theta}{\tan\theta(1-\sin\theta)} = 4$$ for $$0^\circ \leq \theta \leq 360^\circ.$$ 2. **Prove the identity:** Start with the left-hand side (LHS): $$\frac{\cos\theta}{\tan\theta(1-\sin\theta)}.$$ Recall that $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$, so substitute: $$\frac{\cos\theta}{\frac{\sin\theta}{\cos\theta}(1-\sin\theta)} = \frac{\cos\theta}{\frac{\sin\theta(1-\sin\theta)}{\cos\theta}} = \frac{\cos\theta \times \cos\theta}{\sin\theta(1-\sin\theta)} = \frac{\cos^2\theta}{\sin\theta(1-\sin\theta)}.$$ 3. Use the Pythagorean identity $$\cos^2\theta = 1 - \sin^2\theta$$: $$\frac{1 - \sin^2\theta}{\sin\theta(1-\sin\theta)} = \frac{(1 - \sin\theta)(1 + \sin\theta)}{\sin\theta(1-\sin\theta)}.$$ 4. Cancel the common factor $$1 - \sin\theta$$ (assuming $$\sin\theta \neq 1$$): $$\frac{1 + \sin\theta}{\sin\theta}.$$ 5. Rewrite the expression: $$\frac{1}{\sin\theta} + \frac{\sin\theta}{\sin\theta} = \csc\theta + 1.$$ 6. Since $$\csc\theta = \frac{1}{\sin\theta}$$, the expression simplifies to: $$1 + \sin\theta,$$ which matches the right-hand side (RHS). Thus, the identity is proven. 7. **Solve the equation:** Given: $$\frac{\cos\theta}{\tan\theta(1-\sin\theta)} = 4.$$ From the identity, this equals: $$1 + \sin\theta = 4.$$ 8. Solve for $$\sin\theta$$: $$\sin\theta = 4 - 1 = 3.$$ 9. Since $$\sin\theta$$ must be between $$-1$$ and $$1$$, $$\sin\theta = 3$$ is impossible. 10. Check the domain restrictions: The original expression is undefined if $$\tan\theta = 0$$ or $$1 - \sin\theta = 0$$ (i.e., $$\sin\theta = 1$$). 11. Therefore, there is **no solution** to the equation $$\frac{\cos\theta}{\tan\theta(1-\sin\theta)} = 4$$ for $$0^\circ \leq \theta \leq 360^\circ$$. **Final answers:** (i) Identity proven: $$\frac{\cos\theta}{\tan\theta(1-\sin\theta)} \equiv 1 + \sin\theta.$$ (ii) No solutions for $$\theta$$ in $$[0^\circ, 360^\circ]$$ satisfy $$\frac{\cos\theta}{\tan\theta(1-\sin\theta)} = 4.$$