How do I simplify sin((tan^(-1))x)?

$\setminus \frac{x}{\setminus \sqrt{1 + {x}^{2}}}$

Explanation:

Let

$\setminus {\tan}^{- 1} x = \setminus \theta \setminus \quad \left(\setminus \forall \setminus \setminus - \setminus \frac{\pi}{2} \setminus \le \setminus \theta \setminus \le \setminus \frac{\pi}{2}\right)$

$\setminus \tan \setminus \theta = x$

$\setminus \therefore \setminus \sin \setminus \theta = \setminus \frac{\setminus \tan \setminus \theta}{\setminus \sec \setminus \theta}$

$\setminus \sin \setminus \theta = \setminus \frac{\setminus \tan \setminus \theta}{\setminus \sqrt{1 + \setminus {\tan}^{2} \setminus \theta}}$

$\setminus \sin \left(\setminus {\tan}^{- 1} x\right) = \setminus \frac{x}{\setminus \sqrt{1 + {x}^{2}}}$