Question

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

localid="1646686875779" $\sin \theta =\frac{12}{13},\theta insecondquadrant.$

Short answer

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

Step-by-step solution

Step 1. Given information

Here $\sin \theta =\frac{12}{13},\theta insecondquadrant.$is given.

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

. Description of  finding the value of  cos θ 

According to the question$$\begin{gathered} \sin \theta =\frac{12}{13}where\theta lieson2ndquadrant. \\ Weknowthat{\sin }^2\theta +{\cos }^2\theta =1 \\ \cos \theta =\pm \sqrt{1-{\sin }^2\theta }and\sin ce\sin \theta ison2ndquadrantallothertrignometricratioswillbenegative. \\ hence\cos \theta =-\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-\frac{{12}^2}{{13}^2}} \\ =-\sqrt{1-\frac{144}{169}} \\ =-\sqrt{\frac{25}{169}} \\ =-\frac{5}{13} \end{gathered}$$

Step 3. Describing about find the values of cscθ,secθ

Here

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

Step 4.Describing about find the values of tanθ and cotθ

We know that

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