Answer
$\dfrac{3 \sqrt 2}{2} - \dfrac{ 4\sqrt 3}{3}$
Work Step by Step
We know these trigonometric values
$\sin 45^{\circ}=\sin \dfrac{\pi}{4} =\dfrac{ \sqrt 2}{2}$ and $\tan \dfrac{\pi}{6}=\dfrac{\sqrt 3}{3}$
We evaluate the given expression to obtain: $3 \sin 45^{\circ}-4 \tan \dfrac{\pi}{6} =3 \sin \dfrac{\pi}{4}-4 \tan \dfrac{\pi}{6}$
or, $=\dfrac{3 \sqrt 2}{2} - \dfrac{ 4\sqrt 3}{3}$
