Question

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

$\sin \theta =\frac{5}{13},{90}^o<\theta <{180}^o$

Short answer

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

Step-by-step solution

Step 1. Given information

Here $\sin \theta =\frac{5}{13},{90}^o<\theta <{180}^o$are given.

We have to find the exact value of each of the remaining trigonometric functions.

Step 2.Description of finding the value cosθ.

In this question$$\begin{gathered} \sin \theta =\frac{5}{13} \\ Weknowthat{\sin }^2\theta +{\cos }^2\theta =1 \\ \cos \theta =\pm \sqrt{1-{\sin }^2\theta },here\theta lieson{90}^o<\theta <{180}^{o}so\sin \theta isonsecondquadrantallotherexcept\sin \theta andcsc\theta arewillbenegative. \\ hence\cos \theta =-\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-\frac{5^2}{{13}^2}} \\ =-\sqrt{1-\frac{25}{169}} \\ =-\frac{12}{13} \end{gathered}$$

Step 3. finding the value of secθ  and cscθ

Here

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

Step 4. Finding the value for tanθ and cotθ

We know that

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