Answer
$-1$
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 ${{\cos }^{-1}}\left( -\frac{\sqrt{3}}{2} \right)=\alpha $
$\begin{align}
& \frac{-\sqrt{3}}{2}=\cos \alpha \\
& \cos \alpha =\frac{-\sqrt{3}}{2} \\
& =\frac{B}{H}
\end{align}$
By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}=\text{ }{{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{P}^{2}}={{H}^{2}}-{{B}^{2}} \\
& {{P}^{2}}=4-3 \\
& =1
\end{align}$
Now, taking the square root:
$\begin{align}
& P=\sqrt{1} \\
& =1
\end{align}$
Then,
$\begin{align}
& \,\tan \alpha =\frac{P}{B} \\
& =-\frac{1}{\sqrt{3}}\,
\end{align}$
$\begin{align}
& \,\sin \alpha =\frac{\text{P}}{\text{H}} \\
& =\frac{1}{2}
\end{align}$
Step 2: Similarly, ${{\sin }^{-1}}\left( -\frac{1}{2} \right)=\beta $ this implies,
$\begin{align}
& -\frac{1}{2}=\sin \beta \\
& \sin \beta =-\frac{1}{2} \\
& =\frac{P}{\text{H}}
\end{align}$
By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{\text{B}}^{2}}={{\text{H}}^{2}}\text{-}{{\text{P}}^{2}} \\
& {{B}^{2}}={{4}^{2}}-{{1}^{2}} \\
& {{B}^{2}}=3
\end{align}$
Now, taking the square root:
$B=\sqrt{3}$
$\begin{align}
& \cos \beta =\frac{B}{H} \\
& =\frac{\sqrt{3}}{2}
\end{align}$
Step 3: Using steps 1 and 2, $\cos \left[ {{\cos }^{-1}}\left( -\frac{\sqrt{3}}{2} \right)-{{\sin }^{-1}}\left( -\frac{1}{2} \right) \right]$ can be written as:
$\cos \left[ {{\cos }^{-1}}\left( -\frac{\sqrt{3}}{2} \right)-{{\sin }^{-1}}\left( -\frac{1}{2} \right) \right]=\cos \left( \alpha -\beta \right)$
By using the trigonometric identity,
$\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta $
From the above steps, $\sin \alpha =\frac{1}{2}$
$\cos \alpha =-\frac{\sqrt{3}}{2}$ $\cos \beta =\frac{\sqrt{3}}{2}$ $\sin \beta =-\frac{1}{2}$
$\begin{align}
& \,\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta \\
& =\left( -\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2} \right)+\left( \frac{1}{2}\times -\frac{1}{2} \right) \\
& =-\frac{3}{4}-\frac{1}{4} \\
& =\frac{-4}{4}=-1
\end{align}$
