How do you verify (1+sinx+cosx)^2 = 2(1+sinx)(1+cosx)?

You will be squaring a quantity that contains both ##sinx## and ##cosx## so you will need the identity ##sin^2x+cos^2x = 1##. I would start by multiplying both of these out.
##(1+sinx+cosx)^2 = 1 + sinx + cosx + sinx + sin^2x + sinxcosx + cosx + sinxcosx + cos^2x##
##= 1 + 2sinx + 2cosx + sin^2x + cos^2x + 2sinxcosx##
while
##2(1+sinx)(1+cosx) = 2(1+cosx+sinx+sinxcosx) = 2+2cosx+2sinx+2sinxcosx##
Compare:
##1 + sin^2x + cos^2x + cancel(2sinx + 2cosx + 2sinxcosx) = 2+cancel(2cosx+2sinx+2sinxcosx)##
##1 + sin^2x + cos^2x = 2##
##1 + 1 = 2##
##2 = 2##
They’re equal.