Question

a²sin2π+b²tg0+2ab cosπ+b²sin(-π)​

Answer 1

$\sin 2\pi =0, \ \ \mathrm{tg}0=0, \ \ \cos \pi =-1, \ \ \sin (-\pi) =0$

$a^2\sin 2\pi+b^2\mathrm{tg}0+2ab\cos \pi+b^2\sin(-\pi)\\0+0-2ab+0=-2ab$