Chapter 5.2, Problem 94E

To determine

To simplify the expression:

  $\sqrt{u^2-a^2}$

Using substitution: $u=a\sec \theta$ , and $0<\theta <\frac{\pi }{2},a>0$

Answer to Problem 94E

  $\sqrt{u^2-a^2}=a\tan \theta$

Explanation of Solution

Given:

  $\sqrt{u^2-a^2}$

Concept Used:

The trigonometric identity:

• ${\sec }^2\theta -{\tan }^2\theta =1$

Calculation:

In order to solve the expression

  $\sqrt{u^2-a^2}$

Substitute the value of $u=a\sec \theta$ in the expression as,

  $\begin{array}{c}\sqrt{u^2-a^2}=\sqrt{{\left(a\sec \theta \right)}^2-a^2}\\ =\sqrt{a^2{\sec }^2\theta -a^2}\\ =\sqrt{a^2\left({\sec }^2\theta -1\right)}\\ =\sqrt{a^2{\tan }^2\theta },\text{ using }\left({\sec }^2\theta -{\tan }^2\theta =1\right)\\ =a\tan \theta \end{array}$

Thus, putting $u=a\sec \theta$ gives:

  $\sqrt{u^2-a^2}=a\tan \theta$