Chapter 14, Problem 15PT

To determine

To find :The value of $\text{tan}\frac{\pi }{8}$

Answer to Problem 15PT

  $B.\sqrt{2}-1$

Explanation of Solution

Given information :

We know that, the given expression is $\tan \frac{\pi }{8}$

Calculation :

The given equation is

  $\begin{array}{l}\frac{\pi }{8}=\frac{1}{8}\times 180°=22\frac{1}{2}°,\\ Letx=\frac{\pi }{4}\Rightarrow \frac{x}{2}=\frac{\pi }{8}\\ \Rightarrow \tan x\Rightarrow \tan x=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}\\ \Rightarrow \frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}=1\\ \Rightarrow {\tan }^2\left(\frac{x}{2}\right)+2\tan \left(\frac{x}{2}\right)-1=0\\ \Rightarrow \tan \left(\frac{x}{2}\right)=\frac{-2\pm \sqrt{4+4}}{2}=-1\pm \sqrt{2}\\ \text{Now, }\frac{x}{2}=22\frac{1}{2}°\text{ lies in 1st quadrant}\\ \Rightarrow \tan \left(\frac{x}{2}\right)\text{ is positive}\text{.}\\ \text{So, tan }\left(\frac{\text{x}}{\text{2}}\right)=\sqrt{2}-1\end{array}$