Let \(x = \tan \theta\). Then, \(\theta = \tan^{-1} x\).

Substitute \(x = \tan \theta\) into the given expression:

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

Since \(1 + \tan^2 \theta = \sec^2 \theta\), we have:

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

Now, \(\frac{\tan \theta}{\sec \theta} = \frac{\sin \theta / \cos \theta}{1 / \cos \theta} = \sin \theta\).

So, \(\sin^{-1}(\sin \theta) = \theta\).

Since \(\theta = \tan^{-1} x\), the simplified expression is \(\tan^{-1} x\).