Question

The value $\frac{4+\sqrt{13}}{4-\sqrt{13}}-\frac{4-\sqrt{13}}{4+\sqrt{13}}$ is $\frac{16\sqrt{13}}{m}$, then $m$ is

Answer 1

Taking common\quad denominator

$\frac{\left[\right(\sqrt{3}+4\left)\right(\sqrt{13}+4)+(-\sqrt{13}+4\left)\right(-\sqrt{3}+4\left)\right]}{\left[\right(-\sqrt{3}+4)\times (\sqrt{13}+4\left)\right]}$

$=$ $\frac{16\sqrt{13}}{3}$

On comparing this value with $\frac{16\sqrt{13}}{m}$ we have $m=3$