1. **State the problem:** Simplify and verify the equation $$\frac{1}{1+\cos \theta} + \frac{1+\cos \theta}{\sin x} = 2 \csc x$$. 2. **Recall formulas and identities:** - The cosecant function is defined as $$\csc x = \frac{1}{\sin x}$$. - We will try to combine the left side into a single fraction and simplify. 3. **Rewrite the left side with a common denominator:** $$\frac{1}{1+\cos \theta} + \frac{1+\cos \theta}{\sin x} = \frac{\sin x}{(1+\cos \theta)\sin x} + \frac{(1+\cos \theta)^2}{(1+\cos \theta)\sin x}$$ 4. **Combine the fractions:** $$= \frac{\sin x + (1+\cos \theta)^2}{(1+\cos \theta)\sin x}$$ 5. **Expand the numerator:** $$ (1+\cos \theta)^2 = 1 + 2\cos \theta + \cos^2 \theta $$ So numerator becomes: $$ \sin x + 1 + 2\cos \theta + \cos^2 \theta $$ 6. **At this point, the equation involves both $\theta$ and $x$; to verify the equality, assume $\theta = x$ for simplification:** 7. **Substitute $\theta = x$ and rewrite numerator:** $$ \sin x + 1 + 2\cos x + \cos^2 x $$ 8. **Use Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to rewrite $\cos^2 x = 1 - \sin^2 x$:** $$ \sin x + 1 + 2\cos x + 1 - \sin^2 x = \sin x + 2 + 2\cos x - \sin^2 x $$ 9. **The denominator is:** $$ (1+\cos x) \sin x $$ 10. **Rewrite the entire left side:** $$ \frac{\sin x + 2 + 2\cos x - \sin^2 x}{(1+\cos x) \sin x} $$ 11. **Check if this equals the right side:** $$ 2 \csc x = \frac{2}{\sin x} $$ 12. **Multiply both sides by $(1+\cos x) \sin x$ to clear denominators:** $$ \sin x + 2 + 2\cos x - \sin^2 x = 2 (1+\cos x) $$ 13. **Expand right side:** $$ 2 + 2\cos x $$ 14. **Rewrite left side grouping terms:** $$ \sin x - \sin^2 x + 2 + 2\cos x = 2 + 2\cos x $$ 15. **Subtract $2 + 2\cos x$ from both sides:** $$ \sin x - \sin^2 x = 0 $$ 16. **Factor left side:** $$ \sin x (1 - \sin x) = 0 $$ 17. **Solutions:** $$ \sin x = 0 \quad \text{or} \quad \sin x = 1 $$ 18. **Conclusion:** The original equation holds true only for specific values of $x$ where $\sin x = 0$ or $\sin x = 1$. **Final answer:** The equation $$\frac{1}{1+\cos \theta} + \frac{1+\cos \theta}{\sin x} = 2 \csc x$$ is true if $\theta = x$ and $\sin x = 0$ or $\sin x = 1$.