Q.43 - Precalculus Enhanced with Graphing Utilities

Question

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

$\sin \theta =\frac{12}{13} ,\theta in second quadrant.$

Step-by-Step Solution

Answer

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

Step 1. Given information

Here $\sin \theta =\frac{12}{13} ,\theta in second quadrant.$is given.

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

Step 2 . Description of finding the value of cos   θ  

According to the question $$\begin{gathered} \sin \theta =\frac{12}{13} where \theta lies on 2nd quadrant. \\ We know that {\sin }^2\theta +{\cos }^2\theta =1 \\ \cos \theta =\pm \sqrt{1-{\sin }^2\theta } and \sin ce \sin \theta is on 2nd quadrant all other trignometric ratios will be negative. \\ hence \cos \theta =-\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-\frac{{12}^2}{{13}^2}} \\ =-\sqrt{1-\frac{144}{169}} \\ =-\sqrt{\frac{25}{169}} \\ =-\frac{5}{13} \end{gathered}$$

Step 3. Describing about find the values of c s c θ , s e c θ

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 tan θ   a n d   c o t θ

We know that

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