\( \frac { \sin \theta } { \tan \theta } = \sqrt { 1 - \sin ^ { 2 } \theta } = 0 \)

Let's break down this equation! The first part, \( \frac{\sin \theta}{\tan \theta} = \sqrt{1 - \sin^2 \theta} \), simplifies to \( \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta \). So when all is said and done, this means your equation is essentially telling us \( \cos \theta = 0 \). This occurs at \( \theta = \frac{\pi}{2} + n\pi \) for any integer \( n \). Now, for the second part of your equation, it states that \( \sqrt{1 - \sin^2 \theta} = 0 \). This part means \( 1 - \sin^2 \theta \) must equal zero, which leads to \( \sin \theta = \pm 1 \). This happens at \( \theta = \frac{\pi}{2} + 2n\pi \) for any integer \( n \). So, bringing the two parts together, the dominant solution is \( \theta = \frac{\pi}{2} \, (90^\circ)\) where both conditions meet perfectly!