2022-11-23

How do you find the exact value of $\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(-\frac{3}{5}\right)\right)$?

Neil Short

Expert

Let $\theta ={\mathrm{tan}}^{-1}\left(-\frac{3}{5}\right)$
then $\mathrm{tan}\theta =-\frac{3}{5}$ and
${\mathrm{sec}}^{2}\theta =1+{\mathrm{tan}}^{2}\theta =1+\frac{9}{25}=\frac{34}{25}$
As $\mathrm{tan}\theta$ is negative, as range of ${\mathrm{tan}}^{-1}$ is $-\frac{\pi }{2}<\theta <\frac{\pi }{2},\theta$ lies in Q4 and $\mathrm{sec}\theta$ is
positive and $\mathrm{sec}\theta =\frac{\sqrt{34}}{5}$
and hence $\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(-\frac{3}{5}\right)\right)$
$=\mathrm{sec}\theta$
$=\frac{\sqrt{34}}{5}$

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