1. **State the problem:** Solve the equation $$\csc\theta - 1 = \frac{\cot^2\theta}{\csc\theta + 1}$$ for $\theta$. 2. **Recall identities:** - $\csc\theta = \frac{1}{\sin\theta}$ - $\cot\theta = \frac{\cos\theta}{\sin\theta}$ - $\cot^2\theta = \csc^2\theta - 1$ 3. **Rewrite the right side using the identity:** $$\frac{\cot^2\theta}{\csc\theta + 1} = \frac{\csc^2\theta - 1}{\csc\theta + 1}$$ 4. **Factor numerator as difference of squares:** $$\csc^2\theta - 1 = (\csc\theta - 1)(\csc\theta + 1)$$ 5. **Substitute back:** $$\frac{(\csc\theta - 1)(\csc\theta + 1)}{\csc\theta + 1}$$ 6. **Cancel common factor $\csc\theta + 1$ (assuming $\csc\theta \neq -1$):** $$\frac{\cancel{(\csc\theta - 1)}\cancel{(\csc\theta + 1)}}{\cancel{\csc\theta + 1}} = \csc\theta - 1$$ 7. **Now the equation becomes:** $$\csc\theta - 1 = \csc\theta - 1$$ 8. **This is an identity true for all $\theta$ where the original expression is defined, except where $\csc\theta + 1 = 0$, i.e., $\csc\theta = -1$.** 9. **Find where $\csc\theta = -1$: since $\csc\theta = \frac{1}{\sin\theta}$, this means $\sin\theta = -1$.** 10. **$\sin\theta = -1$ at $\theta = \frac{3\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$. These values are excluded because they make the denominator zero.** **Final answer:** The equation holds for all $\theta$ except $\theta = \frac{3\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$.