Dottie Parra

2021-01-04

$\mathrm{sec}\theta =-3,\mathrm{tan}\theta >0$. Find the exact value of the remaining trigonometric functions of
$\theta$.

bahaistag

We have given the value of a trigonometric function $\mathrm{sec}\left(\theta \right)$ and asked the values of other remaining trigonometric functions. So here we will use some standard identities and results in order to determine the values of other trigonometric functions.
${\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta =1$
$\mathrm{cos}\theta =\frac{1}{\mathrm{sec}\theta }$
$\mathrm{sin}\theta =\frac{\mathrm{tan}\theta }{\mathrm{sec}\theta }$
$\mathrm{cos}ec\theta =\frac{1}{\mathrm{sin}\theta }$
$\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }$
Since, $\mathrm{sec}\theta =-3<0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{tan}\theta >0$ then $\theta$ must be in third quadrant in which $\mathrm{sin}\theta ,\mathrm{cos}\theta ,\mathrm{cos}ec\theta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{sec}\theta$ will be negative and rest $\mathrm{tan}\theta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cot}\theta$ will be positive.
Now,
${\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta =1$
${\left(-3\right)}^{2}-{\mathrm{tan}}^{2}\theta =1$
${\mathrm{tan}}^{2}\theta ={\left(-3\right)}^{2}-1$
$\mathrm{tan}\theta =\sqrt{8}$
Also,
$\mathrm{cos}\theta =\frac{1}{\mathrm{sec}\theta }=\frac{1}{-3}=-\frac{1}{3}$
$\mathrm{sin}\theta =\frac{\mathrm{tan}\theta }{\mathrm{sec}\theta }=\frac{\sqrt{8}}{-3}=\frac{-2\sqrt{2}}{3}$
$\mathrm{cos}ec\theta =\frac{1}{\mathrm{sin}\theta }=\frac{1}{\frac{-2\sqrt{2}}{3}}=-\frac{3}{2\sqrt{2}}$
$\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }=\frac{1}{\sqrt{8}}=\frac{1}{2\sqrt{2}}$
Hence, values of other trigonometric functions $\mathrm{sin}\theta ,\mathrm{cos}\theta ,\mathrm{tan}\theta \mathrm{cos}ec\theta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cot}\theta$ will be $\frac{-2\sqrt{2}}{3},-\frac{1}{3},2\sqrt{2},-\frac{3}{2\sqrt{2}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{1}{2\sqrt{2}}$ respectively.

Jeffrey Jordon

Answer is given below (on video)

Jeffrey Jordon