\( \int \begin{array}{l}\text { Prove 1t } \\ \frac{\sin \theta}{\tan \theta}-\sqrt{1-\sin ^{2} \theta}=0\end{array} \)

Let's start with the given equation: \[ \frac{\sin \theta}{\tan \theta}-\sqrt{1-\sin ^{2} \theta}=0. \] First, we recognize that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Substituting this into the equation gives us: \[ \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} - \sqrt{1 - \sin^2 \theta} = 0. \] This simplifies to: \[ \cos \theta - \sqrt{1 - \sin^2 \theta} = 0. \] Using the Pythagorean identity \(1 - \sin^2 \theta = \cos^2 \theta\), we have: \[ \cos \theta - \sqrt{\cos^2 \theta} = 0. \] Since \( \sqrt{\cos^2 \theta} = |\cos \theta|\), we need to consider two cases based on the sign of \(\cos \theta\). 1. **If \(\cos \theta \geq 0\)**: \[ \cos \theta - \cos \theta = 0 \text{ (which is true).} \] 2. **If \(\cos \theta < 0\)**: \[ \cos \theta - (-\cos \theta) = 0 \Rightarrow 2\cos \theta = 0 \Rightarrow \cos \theta = 0 \text{ (which is possible but does not hold for \(\tan \theta\)).} \] Thus, the first case holds where \(\cos \theta \geq 0\), specifically, for \(\theta\) in the range of \(0\) to \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\) (where \(\cos\) is non-negative), proving that: \[ \frac{\sin \theta}{\tan \theta}-\sqrt{1-\sin ^{2} \theta}=0. \] Hence, the original statement is validated and indeed comes out to be true under the specified conditions for \(\theta\).