Answer
$3+\sqrt {2}$
Work Step by Step
RECALL:
$\tan{\theta}=\dfrac{\sin{\theta}}{\cos{\theta}}$
With $\sin{\frac{\pi}{4}}=\frac{\sqrt2}{2}$ and $\cos{\frac{\pi}{4}}=\frac{\sqrt2}{2}$, then
$2\sin \dfrac {\pi }{4}+3\tan \dfrac {\pi }{4}
\\=2\times \left(\dfrac{\sqrt {2}}{2}\right)+3\times \left(\dfrac {\sin \dfrac {\pi }{4}}{\cos \dfrac {\pi }{4}}\right)
\\=\sqrt {2}+3\times \left(\dfrac {\dfrac {\sqrt {2}}{2}}{\dfrac {\sqrt {2}}{2}}\right)
\\=\sqrt {2}+(3\times 1)
\\=\sqrt2+3$
