Chapter 4, Problem 92RE

To determine

To Find: An algebraic expression, which is equivalent to given expression $sec(\arcsin (x))$ .

Answer to Problem 92RE

The algebraic expression of $sec(\arcsin (x))$ is $\frac{1}{\sqrt{1-x^2}}$

Explanation of Solution

Given information:

To draw right angle triangle.

Formula used:

Pythagoras theorem that is $p^2+b^2=h^2$

Where p is perpendicular is base,h is height.

And trigonometric formula,

  $\begin{array}{l}\sec ({\sec }^{-1}(x))=x\\ and\\ si{n}^{-1}x={\cos }^{-1}\sqrt{1-x^2}\\ {\cos }^{-1}(x)={\sec }^{-1}(\frac{1}{x})\end{array}$

Calculation:

Consider the statement “given trigonometric expression”

Recall that by the Pythagoras formula on right angle triangle

  

Therefore, given that perpendicular is $x$

Base is b

Height is $1$

So by Pythagoras formula,

  $\begin{array}{l}\sqrt{p^2+b^2}=h\\ \sqrt{x^2+b^2}=1\\ x^2+b^2=1\\ b^2=1-x^2\\ b=\sqrt{1-x^2}\end{array}$

Where b is base and h is height

  $\begin{array}{l}\cos x=\frac{b}{h}\\ x={\cos }^{-1}\frac{\sqrt{1-x^2}}{1}\end{array}$

Where b is base and h is height

Therefore,

  ${\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}$

Now, recall the trigonometric formula of

  $\begin{array}{l}{\cos }^{-1}x={\sec }^{-1}(\frac{1}{x})\\ therefore\\ {\cos }^{-1}\sqrt{1-x^2}={\sec }^{-1}(\frac{1}{\sqrt{1-x^2}})\end{array}$

Recall the trigonometric formula of $\sec ({\sec }^{-1}x)=x$

So, $\sec ({\sec }^{-1}(\frac{1}{\sqrt{1-x^2}}))=\frac{1}{\sqrt{1-x^2}}$

So, the algebraic value of trigonometric function $\sec ({\arcsin }^{}(x)=\frac{1}{\sqrt{1-x^2}}$ .