Chapter 13.6, Problem 28WE

To determine

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

Answer to Problem 28WE

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

Explanation of Solution

Given:

The given function are $\tan t;\cos t$ .

Calculation:

Write $\tan t=\frac{\sin t}{\cos t}$ .

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

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

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

  $\tan t=\pm \frac{\sqrt{1-{\cos }^{2}x}}{\cos t}$

Therefore, the $\tan t$ in terms of $\cos t$ is as $\tan t=\pm \frac{\sqrt{1-{\cos }^{2}x}}{\cos t}$ .