Tan^-1( x / root of a² - x² )

I think your question is to write the given question in simplest form.
Here is the answer.

$$\begin{gathered} {\tan }^{-1}\left[\frac{\mathrm{x}}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\right] \\ ={\tan }^{-1}\left[\frac{\mathrm{a} \sin \mathrm{\theta }}{\sqrt{{\mathrm{a}}^2 - {\mathrm{a}}^2{\sin }^2\mathrm{\theta }}}\right] \left[\mathrm{put} \mathrm{x} = \mathrm{a} \sin \mathrm{\theta }\right] \\ ={\tan }^{-1}\left[\frac{\mathrm{a} \sin \mathrm{\theta }}{\sqrt{{\mathrm{a}}^2\left(1-{\sin }^2\mathrm{\theta }\right)}}\right] \\ ={\tan }^{-1}\left[\frac{\mathrm{a} \sin \mathrm{\theta }}{\mathrm{a} \cos \mathrm{\theta }}\right] \\ ={\tan }^{-1}\left[\tan \mathrm{\theta }\right] \\ =\mathrm{\theta } \\ ={\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right) \end{gathered}$$