Question

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

$\sin \theta =-\frac{5}{13},where\theta lieson3rdquadrant.$

Short answer

The exact values are$\cos \theta =\frac{-12}{13},csc\theta =-\frac{13}{5},sec\theta =-\frac{13}{12},\tan \theta =\frac{5}{12},cot\theta =\frac{12}{5}$

Step-by-step solution

Step 1. Given information

Here $\sin \theta =-\frac{5}{13}where\theta lieson3rdquadrant$

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

. description of finding the value of cosθ

Here$$\begin{gathered} \sin \theta =-\frac{5}{13} \\ Weknowthat{\sin }^2\theta +{\cos }^2\theta =1 \\ \cos \theta =\pm \sqrt{1-{\sin }^2\theta }and\sin ce\sin \theta isonthirdquadrantallotherexcept\tan \theta andcot\theta arewillbenegative. \\ hence\cos \theta =-\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-\frac{{(-5)}^2}{{13}^2}} \\ =-\sqrt{1-\frac{25}{169}} \\ =-\frac{12}{13} \end{gathered}$$

Step 3. Finding the values of cscθ and secθ

in this question

$$\begin{gathered} \cos \theta =-\frac{12}{13},thensec\theta =\frac{1}{\cos \theta } \\ =\frac{1}{-{\displaystyle \frac{12}{13}}} \\ =-\frac{13}{12} \\ csc\theta =\frac{1}{\sin \theta } \\ =\frac{1}{-{\displaystyle \frac{5}{13}}} \\ =-\frac{13}{5} \end{gathered}$$

Step 4. Finding the values of tan θ, cotθ

Since

$$\begin{gathered} \sin \theta =-\frac{5}{13}and\cos \theta =-\frac{12}{13} \\ \tan \theta =\frac{\sin \theta }{\cos \theta } \\ =\frac{-\frac{5}{13}}{-\frac{12}{13}} \\ =\frac{5}{12} \\ cot\theta =\frac{1}{\tan \theta } \\ =\frac{1}{{\displaystyle \frac{5}{12}}} \\ =\frac{12}{5} \end{gathered}$$