# Question #90557

##### 1 Answer
Aug 22, 2017

$\tan \left(\frac{\pi}{8}\right) = \left(\sqrt{2} - 1\right) .$

#### Explanation:

We have,

${\tan}^{2} \left(\frac{\theta}{2}\right) = {\sin}^{2} \frac{\frac{\theta}{2}}{\cos} ^ 2 \left(\frac{\theta}{2}\right) \times \frac{2}{2} ,$

$= \frac{2 {\sin}^{2} \left(\frac{\theta}{2}\right)}{2 {\cos}^{2} \left(\frac{\theta}{2}\right)} ,$

$= \frac{1 - \cos \theta}{1 + \cos \theta} ,$

$\therefore \tan \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} .$

Letting, $\theta = \frac{\pi}{4} , \text{ so that, } \frac{\theta}{2} = \frac{\pi}{8} ,$ we have,

$\tan \left(\frac{\pi}{8}\right) = \pm \sqrt{\frac{1 - \cos \left(\frac{\pi}{4}\right)}{1 + \cos \left(\frac{\pi}{4}\right)}} ,$

$= \pm \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}} ,$

$= \pm \sqrt{\left\{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right\} \times \left\{\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\right\}} ,$

$= \pm \sqrt{\frac{{\left(\sqrt{2} - 1\right)}^{2}}{2 - 1}} ,$

$\therefore \tan \left(\frac{\pi}{8}\right) = \pm \left(\sqrt{2} - 1\right) .$

As $\left(\frac{\pi}{8}\right) \text{ lies in the } {1}^{s t}$ Quadrant, it is $+ v e .$

$\Rightarrow \tan \left(\frac{\pi}{8}\right) = + \left(\sqrt{2} - 1\right) .$

Enjoy Maths.!