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

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.

• The angle is the inside argument. That is your right triangle. Feb 19 at 4:13
• @DavidP: I would have thought that the argument of the secant would have a natural interpretation as an angle, not the argument of an arctangent. Feb 19 at 4:14
• The argument is $\tan^{-1}(4/3)$, that is the inside argument. Feb 19 at 4:15

\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 $$3,4,5$$
Hope this helps. You need to consider $$\angle \alpha$$.