Q. If $a= 5, b = 13 , c = 12$ in $\Delta ABC$, then $\tan \frac{B}{4}$ is
Solution:
$\tan \frac{B}{4} = \frac{\sin \frac{B}{4}}{\cos \frac{B}{4}}$
$ \tan \frac{B}{4} = \frac{2\sin \frac{B}{4} \cos \frac{B}{4}}{2\cos \frac{B}{4}\cos \frac{B}{4}} $
$= \frac{\sin \frac{B}{2}}{1+ \cos \frac{B}{2}} = \frac{\sin 45^{\circ}}{1+\cos45^{\circ} } \, \, \, \, \, \, \, (\because \, \, \angle B = 90^{\circ})$
$= \frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2} +1}=\sqrt{2} -1$