we that: \( \sin \theta+\sin \theta \cot ^{2} \theta=\csc \)

To solve the equation \( \sin \theta + \sin \theta \cot^2 \theta = \csc \theta \), we start by rewriting the terms in the equation using trigonometric identities. Recall that: - \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) - \( \csc \theta = \frac{1}{\sin \theta} \) Now, we can express \( \cot^2 \theta \) as: \[ \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \] Substituting this into the equation gives: \[ \sin \theta + \sin \theta \cdot \frac{\cos^2 \theta}{\sin^2 \theta} = \csc \theta \] This simplifies to: \[ \sin \theta + \frac{\sin \theta \cos^2 \theta}{\sin^2 \theta} = \csc \theta \] \[ \sin \theta + \frac{\cos^2 \theta}{\sin \theta} = \csc \theta \] Next, we can express \( \csc \theta \) in terms of \( \sin \theta \): \[ \sin \theta + \frac{\cos^2 \theta}{\sin \theta} = \frac{1}{\sin \theta} \] Now, multiply through by \( \sin \theta \) to eliminate the fraction: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we see that this equation holds true for all values of \( \theta \) where \( \sin \theta \neq 0 \). Thus, the original equation \( \sin \theta + \sin \theta \cot^2 \theta = \csc \theta \) is valid for all \( \theta \) except where \( \sin \theta = 0 \) (i.e., \( \theta = n\pi \) for \( n \in \mathbb{Z} \)). In conclusion, the solution to the equation is: \[ \text{Valid for } \theta \neq n\pi, \, n \in \mathbb{Z}. \]