Chapter 4.3, Problem 45E

Solution

To find: the values of $\tan \theta$ and $\sec \theta$ .

  $\tan \theta =-\frac{2\sqrt{21}}{21},\sec \theta =\frac{5\sqrt{21}}{21}$ .

Given:

  $\sin \theta =-\frac{2}{5}$ and $\cos \theta >0$ .

Concept used:

Pythagorean identity:

  ${\sin }^2\theta +{\cos }^2\theta =1$ .

Tangent identity:

  $\tan \theta =\frac{\sin \theta }{\cos \theta }$ .

Reciprocal identity:

  $\sec \theta =\frac{1}{\cos \theta }$

Calculation:

Here, the sine function is negative and the cosine function is positive.

So, $\theta$ lies in quadrant IV.

Now by using the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ , it follows that

  $\begin{array}{c}{\left(-\frac{2}{5}\right)}^2+{\cos }^2\theta =1\\ \frac{4}{25}+{\cos }^2\theta =1\\ {\cos }^2\theta =1-\frac{4}{25}\\ {\cos }^2\theta =\frac{21}{25}\\ \cos \theta =\frac{\sqrt{21}}{5}\end{array}$

Now,

  $\begin{array}{c}\tan \theta =\frac{\sin \theta }{\cos \theta }\\ =\frac{-\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}{\raisebox{1ex}{$\sqrt{21}$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\\ =-\frac{2}{\sqrt{21}}\\ =-\frac{2\sqrt{21}}{21}\end{array}$

And,

  $\begin{array}{c}\sec \theta =\frac{1}{\cos \theta }\\ =\frac{1}{\raisebox{1ex}{$\sqrt{21}$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\\ =\frac{5}{\sqrt{21}}\\ =\frac{5\sqrt{21}}{21}\end{array}$