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:
$$\sec(\sin^{-1}(-5/13)-\tan^{-1}(4/3))$$
 A: If $\displaystyle A=\sin^{-1}\left(-\frac5{13}\right)$ and  $\displaystyle\ B=\tan^{-1}\left(\frac43\right)$
Using Principal values, $\displaystyle -\frac\pi2<A<0$ and  $\displaystyle 0<B<\frac\pi2$
$\displaystyle\implies\sin A=-\frac5{13},\cos A=+\sqrt{1-\left(-\frac5{13}\right)^2}=+\frac{12}{13}$
$\displaystyle\implies\tan B=\frac43,\sec B=+\sqrt{1+\left(\frac43\right)^2}=+\frac53\iff \cos B=\frac35$
$\displaystyle\implies\sin B=\tan B\cdot\cos B=\cdots$
$\displaystyle\implies\cos(A-B)=\cos A\cos B+\sin A\sin B$
Finally, $\displaystyle\sec x=\frac1{\cos x}$
A: Since $\sec(x) = 1/\cos(x)$, let's first consider
$$\cos\left[\sin^{-1}(-5/13) - \tan^{-1}(4/3)\right].$$
Using the addition formula for $\cos$, this is equal to
$$\cos(\sin^{-1}(-5/13)) \cos(\tan^{-1}(4/3)) + \sin(\sin^{-1}(-5/13)) \sin(\tan^{-1}(4/3))$$
Now let's work term-by-term.
The key is to recognise that the arguments of the inverse trig functions are part of Pythagorean triples.
First, $\cos(\sin^{-1}(-5/13))$.
If we write $u = \sin^{-1}(-5/13)$, then $\sin u = -5/13$.
These numbers might remind you of the Pythagorean triple $(5,12,13)$, so if
$$-\sin u = \sin(-u) = \frac{\text{OPP}}{\text{HYP}} = \frac{5}{13},$$
then we must have
$$\cos(-u) = \frac{\text{ADJ}}{\text{HYP}} = \frac{12}{13}.$$
Further, since $\cos(-u) = \cos(u)$ for all real $u$, so our first term is $12/13$.
A similar principle works for our terms with $\tan^{-1}(4/3)$, where we use the Pythagorean triple $(3,4,5)$. Write $w = \tan^{-1}(4/3)$. If $\text{OPP}=4$, $\text{ADJ}=3$ and $\text{HYP} = 5$, then
$$\sin(w) = \frac{4}{5} \text{ and } \cos(w) = \frac{3}{5}.$$
Finally, the $\sin(\sin^{-1}(-5/13))$ term simplifies to $-5/13$ (strictly speaking, this depends on our choice of domain for $\sin^{-1}$).
Plugging this all in, we have
$$\cos\left[\sin^{-1}(-5/13) - \tan^{-1}(4/3)\right]
= \frac{12}{13} \cdot \frac{3}{5} + \frac{-5}{13} \cdot \frac{4}{5}
= \frac{16}{65}.
$$
Then we just flip this fraction to get what you actually want:
$$\sec\left[\sin^{-1}(-5/13) - \tan^{-1}(4/3)\right] = \frac{65}{16}.$$
A: $\sec\left(\sin^{-1}\left(\dfrac{-5}{13}\right)-\tan^{-1}\left(\dfrac{4}{3}\right)\right)$
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}
