Chapter 13.6, Problem 30WE

To determine

To write: The first function in terms of the second.

Answer to Problem 30WE

The $\tan \alpha \sec \alpha$ in terms of $\cos \alpha$ is as $\tan \alpha \sec \alpha =\pm \frac{\sqrt{1-{\cos }^2\alpha }}{{\cos }^2\alpha }$ .

Explanation of Solution

Given:

The given function are $\tan \alpha \sec \alpha ;\cos \alpha$ .

Calculation:

Simplify the expression as follows.

  $\begin{array}{c}\tan \alpha \sec \alpha =\frac{\sin \alpha }{\cos \alpha }\cdot \frac{1}{\cos \alpha }\\ \tan \alpha \sec \alpha =\frac{\sin \alpha }{{\cos }^2\alpha }\end{array}$

Use the identity ${\sin }^{2}t+{\cos }^{2}t=1$ to find $\sin \alpha$ .

  $\begin{array}{c}{\sin }^2\alpha =1-{\cos }^2\alpha \\ \sin \alpha =\pm \sqrt{1-{\cos }^2\alpha }\end{array}$

Substitute $\pm \sqrt{1-{\cos }^2\alpha }$ for $\sin \alpha$ in the equation $\tan \alpha \sec \alpha =\frac{\sin \alpha }{{\cos }^2\alpha }$ .

  $\tan \alpha \sec \alpha =\pm \frac{\sqrt{1-{\cos }^2\alpha }}{{\cos }^2\alpha }$

Therefore, the $\tan \alpha \sec \alpha$ in terms of $\cos \alpha$ is as

  $\tan \alpha \sec \alpha =\pm \frac{\sqrt{1-{\cos }^2\alpha }}{{\cos }^2\alpha }$ .