Answer
The required value is $-\frac{33}{65}$
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}}\frac{4}{3}=\alpha $
$\begin{align}
& \frac{4}{3}=\tan \alpha \\
& \tan \alpha =\frac{4}{3} \\
& =\frac{\text{P}}{\text{B}}
\end{align}$
By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}=\text{ }{{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{\text{H}}^{2}}={{4}^{2}}+{{3}^{2}} \\
& {{\text{H}}^{2}}=16+9 \\
& =25
\end{align}$
Now, taking the square root:
$\begin{align}
& \text{H}=\sqrt{25} \\
& =5
\end{align}$
Then,
$\begin{align}
& \,\cos \alpha =\frac{\text{B}}{\text{H}} \\
& =\frac{3}{5}\,
\end{align}$
$\begin{align}
& \,\sin \alpha =\frac{\text{P}}{\text{H}} \\
& =\frac{4}{5}
\end{align}$
Step 2: Similarly, ${{\cos }^{-1}}\frac{5}{13}=\beta $ this implies,
$\begin{align}
& \frac{5}{13}=\cos \beta \\
& \cos \beta =\frac{5}{13} \\
& =\frac{\text{B}}{\text{H}}
\end{align}$
By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{13}^{2}}={{\text{P}}^{2}}+{{5}^{2}} \\
& {{\text{P}}^{2}}=169-25 \\
& =144
\end{align}$
Now, taking the square root:
$\begin{align}
& \text{P}=\sqrt{144} \\
& =12
\end{align}$
$\begin{align}
& \sin \beta =\frac{\text{P}}{\text{H}} \\
& =\frac{12}{13}
\end{align}$
Step 3: Using steps 1 and 2, $\cos \left( ta{{n}^{-1}}\frac{4}{3}+{{\cos }^{-1}}\frac{5}{13} \right)$ can be written as:
$\cos \left( ta{{n}^{-1}}\frac{4}{3}+{{\cos }^{-1}}\frac{5}{13} \right)=\cos \left( \alpha +\beta \right)$
By using the trigonometric identity,
$\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta $
From the above steps, $\sin \alpha =\frac{4}{5}$
$\cos \alpha =\frac{3}{5}$ $\cos \beta =\frac{5}{13}$ $\sin \beta =\frac{12}{13}$
$\begin{align}
& \,\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\
& =\left( \frac{3}{5}\times \frac{5}{13} \right)-\left( \frac{4}{5}\times \frac{12}{13} \right) \\
& =\frac{15}{65}-\frac{48}{65} \\
& =-\frac{33}{65}
\end{align}$