Chapter 9.4, Problem 32E

Solution

To rewrite: the given expression.

  $\csc \left({\sin }^{-1}x\right)=\frac{1}{x}$ .

Given:

  $\csc \left({\sin }^{-1}x\right)$ .

Calculation:

Let ${\sin }^{-1}x=\theta$ . Then,

  $x=\sin \theta$

By taking the reciprocals on both sides, it follows that

  $\begin{array}{c}\frac{1}{x}=\frac{1}{\sin \theta }\\ \frac{1}{x}=\csc \theta \\ \theta ={\csc }^{-1}\left(\frac{1}{x}\right)\\ {\sin }^{-1}x={\csc }^{-1}\left(\frac{1}{x}\right)\end{array}$

Therefore,

  $\begin{array}{c}\csc \left({\sin }^{-1}x\right)=\csc \left({\csc }^{-1}\frac{1}{x}\right)\\ =\frac{1}{x}\end{array}$