Chapter 3.6, Problem 2E

  $\left(a\right)$

To determine

To calculate: The value of trigonometric expression ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ .

Answer to Problem 2E

The value of trigonometric expression ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is $\frac{\pi }{6}$ .

Explanation of Solution

Given information:

The trigonometric expression ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ .

Formula used:

Inverse tangent trigonometric function is expressed as ${\tan }^{-1}\left(x\right)=y$ which is equivalent to $x=\tan y$ where $-\frac{\pi }{2}<y<\frac{\pi }{2}$ .

Calculation:

Consider trigonometric expression ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ .

Recall that the tangent trigonometric function is expressed as ${\tan }^{-1}\left(x\right)=y$ which is equivalent to $x=\tan y$ where $-\frac{\pi }{2}<y<\frac{\pi }{2}$ .

Recall that value of tangent function at an angle $\frac{\pi }{6}$ is $\frac{1}{\sqrt{3}}$ .

Also the angle $\frac{\pi }{6}$ lies between $-\frac{\pi }{2}\text{ and }\frac{\pi }{2}$ .

That is, $\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$ .

Take sine function to the right hand side,

  ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$ .

Thus, the value oftrigonometric expression ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is $\frac{\pi }{6}$ .

  $\left(b\right)$

To determine

To calculate: The value of trigonometric expression ${\sec }^{-1}\left(2\right)$ .

Answer to Problem 2E

The value of trigonometric expression ${\sec }^{-1}\left(2\right)$ is $\frac{\pi }{3}$ .

Explanation of Solution

Given information:

The trigonometric expression ${\sec }^{-1}\left(2\right)$ .

Formula used:

Inverse secant trigonometric function is expressed as ${\sec }^{-1}\left(x\right)=y$ which is equivalent to $x=\sec y$ where $-\frac{\pi }{2}<y<\frac{\pi }{2}$ .

Calculation:

Consider trigonometric expression ${\sec }^{-1}\left(2\right)$ .

Recall that the inverse secant trigonometric function is expressed as ${\sec }^{-1}\left(x\right)=y$ which is equivalent to $x=\sec y$ where $-\frac{\pi }{2}<y<\frac{\pi }{2}$ .

Recall that value of secant function at an angle $\frac{\pi }{3}$ is $2$ .

Also the angle $\frac{\pi }{3}$ lies between $-\frac{\pi }{2}\text{ and }\frac{\pi }{2}$ .

That is, $\sec \left(\frac{\pi }{3}\right)=2$ .

Take cosine function to the right hand side,

  ${\sec }^{-1}\left(2\right)=\frac{\pi }{3}$ .

Thus, the value of trigonometric expression ${\sec }^{-1}\left(2\right)$ is $\frac{\pi }{3}$ .