Answer
The required value is $-\frac{24}{25}$
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}}\frac{3}{5}=\alpha $
$\begin{align}
& \frac{3}{5}=\sin \alpha \\
& \sin \alpha =\frac{3}{5}=\frac{\text{P}}{\text{H}} \\
\end{align}$
By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}=\text{ }{{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{5}^{2}}={{\text{3}}^{2}}+{{\text{B}}^{2}} \\
& {{\text{B}}^{2}}=25-9 \\
& =16
\end{align}$
Now, taking the square root of both sides
$\begin{align}
& \text{B}=\sqrt{16} \\
& =4
\end{align}$
$\text{So,}\,\cos \alpha =\frac{\text{B}}{\text{H}}=\frac{4}{5}$
Step 2: Similarly, ${{\cos }^{-1}}\left( -\frac{4}{5} \right)=\beta $
$\begin{align}
& \text{This}\,\text{implies,}\,-\frac{4}{5}=\cos \beta \\
& \cos \beta =-\frac{4}{5}=\frac{\text{B}}{\text{H}} \\
\end{align}$
By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}=\text{ }{{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{5}^{2}}={{\text{P}}^{2}}+{{\left( -4 \right)}^{2}} \\
& {{\text{P}}^{2}}=25-16 \\
& =9
\end{align}$
Now, taking the square root:
$\begin{align}
& \text{P}=\sqrt{9} \\
& =3
\end{align}$
$\text{So,}\,\sin \beta =\frac{\text{P}}{\text{H}}=\frac{3}{5}$
Step 3: Using steps 1 and 2, $\sin \left[ {{\sin }^{-1}}\frac{3}{5}-{{\cos }^{-1}}\left( -\frac{4}{5} \right) \right]$ can be written as:
$\sin \left[ {{\sin }^{-1}}\frac{3}{5}-{{\cos }^{-1}}\left( -\frac{4}{5} \right) \right]=\sin \left( \alpha -\beta \right)$
By using the trigonometric identity,
$\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $
From above steps, $\sin \alpha =\frac{3}{5}$ $\cos \beta =-\frac{4}{5}$ $\sin \beta =\frac{3}{5}$
$\begin{align}
& \text{Thus,}\,\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\
& =\left( \frac{3}{5}\times \left( -\frac{4}{5} \right) \right)-\left( \frac{4}{5}\times \frac{3}{5} \right) \\
& =-\frac{12}{25}-\frac{12}{25} \\
& =-\frac{24}{25}
\end{align}$
