How to solve this trig problem? $\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$ Basic trig problem my brother ask me, but I don't know how to do it:
$$\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$$

$\sec\left(\sin^{-1}\left(\dfrac{-5}{13}\right)-\tan^{-1}\left(\dfrac{4}{3}\right)\right)$
Let $\alpha = \sin^{-1}\left(\dfrac{-5}{13}\right)$
Then $\alpha$ is the angle in the fourth quadrant that corresponds to the point
$(x,y) = (12,-5)$ with amplitude $r = 13.$
So $\cos(\alpha) = \dfrac xr = \dfrac{12}{13}$ and
   $\sin(\alpha) = \dfrac yr = \dfrac{-5}{13}$
Let $\beta = \tan^{-1}\left(\dfrac{4}{3}\right)$
Then $\beta$ is the angle in the first quadrant that corresponds to the point $(x,y) = (3,4)$ with amplitude $r = 5.$
So $\cos(\beta) = \dfrac xr = \dfrac 35$ and
   $\sin(\beta) = \dfrac yr = \dfrac 45$
So
\begin{align}
   \sec\left(\sin^{-1}\left(\dfrac{-5}{13}\right)-
             \tan^{-1}\left(\dfrac{4}{3}\right)\right)
    &= \sec(\alpha - \beta)\\
    &= \dfrac{1}{\cos(\alpha - \beta)}\\
    &= \dfrac{1}{\cos \alpha \; \cos \beta + \sin \alpha \; \sin \beta}\\
    &= \dfrac{1}{\dfrac{12}{13} \cdot \dfrac{3}{5} +
       \dfrac{-5}{13} \cdot \dfrac 45}\\
    &= \dfrac{65}{16}
\end{align}
