What will be the value of (tan 45°- tan 15°)/(1+tan 45° tan 15°)?

$\frac{tan 45°- tan 15°}{1+tan 45° tan 15°}$

= tan(45 - 15)° = tan30° = $\frac{1}{\sqrt{3}}$

Since, $\frac{tan A- tan B}{1+tan A tan B}$  =tan(A - B)