Question

If $a=5,b=13,c=12$ in $\Delta ABC$, then $\tan \frac{B}{4}$ is

• A. $\sqrt{3}-1$
• B. $\sqrt{3}+1$
• C. $\sqrt{2}+1$
• D. $\sqrt{2}-1$

Answer 1

Answer: (D)

$\tan \frac{B}{4}=\frac{\sin \frac{B}{4}}{\cos \frac{B}{4}}$
$\tan \frac{B}{4}=\frac{2\sin \frac{B}{4}\cos \frac{B}{4}}{2\cos \frac{B}{4}\cos \frac{B}{4}}$
$=\frac{\sin \frac{B}{2}}{1+\cos \frac{B}{2}}=\frac{\sin 45^{\circ }}{1+\cos 45^{\circ }}(\because \angle B=90^{\circ })$
$=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$