Creating a relevant right triangle to evaluate $\sec\left(\arctan\frac{4}{3}\right)$

Question

I am trying to solve the following:

$$\sec\left(\arctan\left(\frac{4}{3}\right)\right)$$

The problem tells me to use a relevant right triangle, but I am curious as to if I need to create a right triangle for the inside ($\arctan(\frac{4}{3}))$ or the outside ($\sec$)?

Thank you.

Answer 1

$$\begin{align} &\theta =\mathrm{arc}\tan \displaystyle\frac{4}{3} &\\ &\tan \theta =\displaystyle\frac{4}{3} &\\ &\sec ^2\theta =1+\tan ^2\theta =\displaystyle\frac{25}{9} &\\ &\sec \theta =\displaystyle\frac{5}{3}\\ \end{align}$$ since $\theta\in(0,\pi/2)$,so wo ignore the negative value

from the value of $tan$ you can see this is a right triangle with length $$3,4,5$$

Answer 2

Hope this helps. You need to consider $\angle \alpha$.