Chapter 13.6, Problem 29WE

To determine

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

Answer to Problem 29WE

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

Explanation of Solution

Given:

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

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 ${\cos }^2\alpha$ .

  ${\cos }^2\alpha =1-{\sin }^2\alpha$

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

  $\tan \alpha \sec \alpha =\frac{\sin \alpha }{1-{\sin }^2\alpha }$

Therefore, the $\tan \alpha \sec \alpha$ in terms of $\sin \alpha$ is as $\tan \alpha \sec \alpha =\frac{\sin \alpha }{1-{\sin }^2\alpha }$ .