Chapter 12, Problem 7CR

To determine

Find the value of $\tan \theta$ such that $\cos \theta =\frac{8}{17}\text{ and }\sin \theta >0$

Answer to Problem 7CR

The value of $\tan \theta$ is $\frac{15}{8}$ , so option (a) is correct.

Explanation of Solution

Given information: The given trigonometric function is,

  $\cos \theta =\frac{8}{17}\text{ and }\sin \theta >0\mathrm{....}\left(1\right)$

Calculation: From (1) the cosine and sine functions are positive which means the angle $\left(\theta \right)$ lie in 1^st quadrant. Therefore the tangent function evaluated in 1^st quadrant will also be positive. Therefore the value of $\tan \theta$ using (1) is evaluated as,

  $\begin{array}{c}\cos \theta =\frac{8}{17}\text{ and }\sin \theta >0\mathrm{....}\left\{\text{From }\left(1\right)\right\}\\ \sin \theta =\pm \sqrt{1-{\cos }^2\theta }\\ =\pm \sqrt{1-{\left(\frac{8}{17}\right)}^2}\\ =\frac{15}{17}\mathrm{....}\left(\text{Take positive value}\right)\end{array}$

Now evaluate for $\tan \theta$ ,

  $\begin{array}{c}\tan \theta =\frac{\sin \theta }{\cos \theta }\\ =\frac{\left(15/17\right)}{\left(8/17\right)}\\ =\frac{15}{8}\end{array}$

Hence the value of $\tan \theta$ is $\frac{15}{8}$ , so option (a) is correct.