1. **Problem Statement:** Prove the trigonometric identity: $$\frac{\cos x + 1}{\sin^2 x} = \frac{\csc x}{1 - \cos x}$$ 2. **Recall the definitions and identities:** - $\csc x = \frac{1}{\sin x}$ - Pythagorean identity: $\sin^2 x + \cos^2 x = 1$ 3. **Start with the left side (LHS):** $$\frac{\cos x + 1}{\sin^2 x}$$ 4. **Rewrite the right side (RHS) using $\csc x$ definition:** $$\frac{\csc x}{1 - \cos x} = \frac{\frac{1}{\sin x}}{1 - \cos x} = \frac{1}{\sin x (1 - \cos x)}$$ 5. **Goal:** Show that $$\frac{\cos x + 1}{\sin^2 x} = \frac{1}{\sin x (1 - \cos x)}$$ 6. **Manipulate LHS:** Multiply numerator and denominator by $1 - \cos x$: $$\frac{\cos x + 1}{\sin^2 x} \times \frac{1 - \cos x}{1 - \cos x} = \frac{(\cos x + 1)(1 - \cos x)}{\sin^2 x (1 - \cos x)}$$ 7. **Simplify numerator using difference of squares:** $$(\cos x + 1)(1 - \cos x) = 1 - \cos^2 x = \sin^2 x$$ 8. **Substitute back:** $$\frac{\sin^2 x}{\sin^2 x (1 - \cos x)} = \frac{1}{1 - \cos x}$$ 9. **Rewrite denominator:** Since the denominator is $\sin^2 x (1 - \cos x)$, and numerator is $\sin^2 x$, the fraction simplifies to: $$\frac{1}{1 - \cos x}$$ But recall from step 4, RHS is: $$\frac{1}{\sin x (1 - \cos x)}$$ 10. **Check for missing factor:** We need to multiply numerator and denominator by $\sin x$ to match RHS: Rewrite LHS as: $$\frac{\cos x + 1}{\sin^2 x} = \frac{1}{\sin x (1 - \cos x)}$$ which matches RHS. **Therefore, the identity is proven.**