Answer
5
Work Step by Step
$\sin 90^{\circ}=1$
Cosine of all odd multiples of $90^{\circ}$ are 0.
$\implies \cos 270^{\circ}=0$
$\tan 360^{\circ}=\tan 0^{\circ}=0$
Then,
$5\sin^{2} 90^{\circ}+2\cos^{2}270^{\circ}-\tan 360^{\circ}=$
$5(1)^{2}+2(0)^{2}-0=5$
