Answer
$\dfrac{3\sqrt2}{2}-\dfrac{4\sqrt3}{3}$
Work Step by Step
With $\sin{45^\circ}=\sin{\left(\frac{\pi}{4}\right)}=\frac{\sqrt2}{2}$ and $\tan{\left(\frac{\pi}{6}\right)}=\frac{\sqrt3}{3}$, evaulating the given expression yields:
$3\sin{45^\circ}-4\tan{\frac{\pi}{6}}\\
=3\sin{\left(\dfrac{\pi}{4}\right)}-4\tan{\left(\dfrac{\pi}{6}\right)}\\
=3\left(\dfrac{\sqrt2}{2}\right)-4\left(\dfrac{\sqrt3}{3}\right)\\
=\dfrac{3\sqrt2}{2}-\dfrac{4\sqrt3}{3}$
