Chapter 13.2, Problem 2WE

To determine

To verify: The identity $s<\tan s$ for $0<s<\frac{\pi }{2}$ .

Answer to Problem 2WE

The identity $s<\tan s$ is proved for $0<s<\frac{\pi }{2}$ .

Explanation of Solution

Given information:

The givendiagram is shown in figure (1).

  

Figure (1)

Calculation:

Using the triangle $\Delta \text{OPM}$ and $\Delta \text{OQA}$ are similar

  $\frac{PM}{QA}=\frac{OM}{OA}$

The given arc is a part of a unit circle, the $OA$ is unity.

  $\frac{\sin s}{AQ}=\frac{\cos s}{1}$

Multiply each side of the equation by $\frac{AQ}{\cos s}$ and simplify.

  $\begin{array}{c}\frac{AQ}{\cos s}\left(\frac{\sin s}{AQ}\right)=\frac{AQ}{\cos s}\left(\frac{\cos s}{1}\right)\\ \frac{\sin s}{\cos s}=AQ\\ \tan s=AQ\end{array}$

Calculate are of the triangle $\Delta \text{OQA}$ .

  $\begin{array}{c}Area=\frac{1}{2}\left(OA\right)\left(AQ\right)\\ =\frac{1}{2}\left(1\right)\left(\tan s\right)\\ =\frac{\tan s}{2}\end{array}$

Calculate the area of sector $OAP$ .

  $\begin{array}{c}Area=\frac{1}{2}\left(\text{radius}\right)\left(\text{length}\text{of}\text{arc}\right)\\ =\frac{1}{2}\left(1\right)\left(s\right)\\ =\frac{s}{2}\end{array}$

From figure we can see that the area of sector $OAP$ is less than the area of $\Delta \text{OQA}$ .

  $\frac{s}{2}<\frac{\tan s}{2}$

Multiply each side by $2$ .

  $s<\tan s$

Therefore, the identity $s<\tan s$ is proved for $0<s<\frac{\pi }{2}$ .