What will be the value of $\frac{tan 45°- tan 15°}{1+tan 45° tan 15°}$?

$\sqrt{3}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{\sqrt{3}}$ $\sqrt{2}$

$\frac{1}{\sqrt{3}}$

$\frac{tan 45°- tan 15°}{1+tan 45° tan 15°}$ = tan(45 - 15)° = tan30° = $\frac{1}{\sqrt{3}}$ Since, $\frac{tan A- tan B}{1+tan A tan B}$  =tan(A - B)