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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.

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    $\begingroup$ The angle is the inside argument. That is your right triangle. $\endgroup$
    – David P
    Feb 19 at 4:13
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    $\begingroup$ @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. $\endgroup$
    – joriki
    Feb 19 at 4:14
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    $\begingroup$ The argument is $\tan^{-1}(4/3)$, that is the inside argument. $\endgroup$
    – David P
    Feb 19 at 4:15

2 Answers 2

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$$\begin{align} &\theta =\mathrm{arc}\tan \displaystyle\frac{4}{3} &\\ &\tan \theta =\displaystyle\frac{4}{3} &\\ &\sec ^2\theta =1+\tan ^2\theta =\displaystyle\frac{25}{9} &\\ &\sec \theta =\displaystyle\frac{5}{3}\\ \end{align}$$ 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 $$3,4,5$$

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

enter image description here

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