Answer
The required value of the expression is $\frac{x\sqrt{1-{{y}^{2}}}-y}{\sqrt{{{x}^{2}}+1}}$.
Work Step by Step
Step 1: As, $\sin \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Hypotenuse}\left( \text{H} \right)}$ $\cos \theta =\frac{\text{Base}\left( \text{B} \right)}{\text{Hypotenuse}\left( \text{H} \right)}$ $\tan \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Base}\left( \text{B} \right)}$
Let us suppose ${{\tan }^{-1}}x=\alpha $
$\begin{align}
& x=\tan \alpha \\
& \tan \alpha =\frac{x}{1} \\
& =\frac{\text{P}}{\text{B}}
\end{align}$
By using the Pythagorian Theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{\text{H}}^{2}}={{x}^{2}}+{{1}^{2}} \\
& {{\text{H}}^{2}}={{x}^{2}}+1 \\
& \text{H}=\sqrt{{{x}^{2}}+1} \\
\end{align}$
Therefore,
$\begin{align}
& \sin \alpha =\frac{\text{P}}{\text{H}} \\
& =\frac{x}{\sqrt{{{x}^{2}}+1}}
\end{align}$
Then,
$\begin{align}
& \cos \alpha =\frac{\text{B}}{\text{H}} \\
& =\frac{1}{\sqrt{{{x}^{2}}+1}}
\end{align}$
Step 2: Similarly, ${{\sin }^{-1}}y=\beta $. This implies,
$\begin{align}
& y=\sin \beta \\
& \sin \beta =\frac{y}{1} \\
& =\frac{\text{P}}{\text{H}}
\end{align}$
By using the Pythagorian Theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{1}^{2}}={{y}^{2}}+{{\text{B}}^{2}} \\
& {{\text{B}}^{2}}=1-{{y}^{2}} \\
& \text{B}=\sqrt{1-{{y}^{2}}} \\
\end{align}$
So,
$\begin{align}
& \cos \beta =\frac{\text{B}}{\text{H}} \\
& =\frac{\sqrt{1-{{y}^{2}}}}{1} \\
& =\sqrt{1-{{y}^{2}}}
\end{align}$
Step 3: Using steps 1 and 2, $\sin \left( {{\tan }^{-1}}x-{{\sin }^{-1}}y \right)$ can be written as:
$\sin \left( {{\tan }^{-1}}x-{{\sin }^{-1}}y \right)=\sin \left( \alpha -\beta \right)$
By using the trigonometric identity,
$\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $
From above steps, $\sin \alpha =\frac{x}{\sqrt{{{x}^{2}}+1}}$ $\cos \alpha =\frac{1}{\sqrt{{{x}^{2}}+1}}$ $\cos \beta =\sqrt{1-{{y}^{2}}}$ $\sin \beta =y$
Thus,
$\begin{align}
& \sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\
& =\left( \frac{x}{\sqrt{{{x}^{2}}+1}}\times \sqrt{1-{{y}^{2}}} \right)-\left( \frac{1}{\sqrt{{{x}^{2}}+1}}\times y \right) \\
& =\frac{x\sqrt{1-{{y}^{2}}}}{\sqrt{{{x}^{2}}+1}}-\frac{y}{\sqrt{{{x}^{2}}+1}} \\
& =\frac{x\sqrt{1-{{y}^{2}}}-y}{\sqrt{{{x}^{2}}+1}}
\end{align}$
