Question

Find the exact value of each of the remaining trigonometric functions of $\theta$

$\cos \theta =\frac{-4}{5}in3rdquadrant$

Short answer

the values are$\sin \theta =-\frac{3}{5},sec\theta =-\frac{5}{4},csc\theta =-\frac{5}{3},\tan \theta =\frac{3}{4},cot\theta =\frac{4}{3}$

Step-by-step solution

Step 1. Given information 

Here$\cos \theta =\frac{-4}{5},where\theta liesonthirdqaudrantisgiven$

we have to find exact value of each of the remaining trigonometric functions

Step 2.Description of  finding the value of sinθ

$$\begin{gathered} \cos \theta =-\frac{4}{5} \\ Weknowthat{\sin }^2\theta +{\cos }^2\theta =1 \\ \sin \theta =\pm \sqrt{1-{\cos }^2\theta }and\sin ce\cos \theta isonthirdquadrantallotherexcept\tan \theta andcot\theta arewillbenegative. \\ hence\sin \theta =-\sqrt{1-{\cos }^2\theta } \\ =-\sqrt{1-\frac{{(-4)}^2}{5^2}} \\ =-\sqrt{1-\frac{16}{25}} \\ =-\frac{3}{5} \end{gathered}$$

Step 3. Finding the values of secθ and cscθ

According to the question$$\begin{gathered} \cos \theta =-\frac{4}{5},thensec\theta =\frac{1}{\cos \theta } \\ =\frac{1}{-{\displaystyle \frac{4}{5}}} \\ =-\frac{5}{4} \\ csc\theta =\frac{1}{\sin \theta } \\ =\frac{1}{-{\displaystyle \frac{3}{5}}} \\ =-\frac{5}{3} \end{gathered}$$

Step 4. Finding the value of tanθ and cotθ

since

$$\begin{gathered} \sin \theta =-\frac{3}{5}and\cos \theta =-\frac{4}{5} \\ \tan \theta =\frac{\sin \theta }{\cos \theta } \\ =\frac{-\frac{3}{5}}{-\frac{4}{5}} \\ =\frac{3}{4} \\ cot\theta =\frac{1}{\tan \theta } \\ =\frac{1}{{\displaystyle \frac{3}{4}}} \\ =\frac{4}{3} \end{gathered}$$