Answer
The inverse trigonometric function can be written as $\frac{x\sqrt{{{x}^{2}}-1}}{\sqrt{{{x}^{2}}-1}}$.
Work Step by Step
Let us consider $\theta ={{\sin }^{-1}}\frac{1}{x}$; then multiply on both sides by $\sin $.
Since, $\sin \theta =\frac{1}{x}$
Use the $\sin $ formula in the right triangle $ABC$.
So,
$\sin \theta =\frac{p}{r}$
Use the Pythagorean Theorem to find the base.
$\begin{align}
& {{r}^{2}}={{p}^{2}}+{{b}^{2}} \\
& {{x}^{2}}={{1}^{2}}+{{b}^{2}} \\
& {{b}^{2}}={{x}^{2}}-1 \\
& b=\sqrt{{{x}^{2}}-1}
\end{align}$
Then, use the $\sec $ formula in the above right triangle:
$\begin{align}
& \sec \theta =\frac{r}{b} \\
& =\frac{x}{\sqrt{{{x}^{2}}-1}}
\end{align}$
Then, rationalize the above expression as:
$\frac{x}{\sqrt{{{x}^{2}}-1}}\times \frac{\sqrt{{{x}^{2}}-1}}{\sqrt{{{x}^{2}}-1}}=\frac{x\sqrt{{{x}^{2}}-1}}{{{x}^{2}}-1}$
Hence, the inverse trigonometric function can be written as $\frac{x\sqrt{{{x}^{2}}-1}}{\sqrt{{{x}^{2}}-1}}$.
