Chapter 13.6, Problem 27WE

To determine

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

Answer to Problem 27WE

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

Explanation of Solution

Given:

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

Calculation:

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

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

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

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

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

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