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