∵sinxcosx=1/2,
∴(sinx+cosx)²
=sin²x+cos²x+2sinxcosx
=1+2×1/2
=2
∵x∈[0,π/2],
∴sinx+cosx>0
∴sinx+cosx=√2
∴1/(1+sinx)+1/(1+cosx)
=[(1+cosx)+(1+sinx)]/[(1+sinx)(1+cosx)]
=(2+sinx+cosx)/[1+(sinx+cosx)+sinxcosx]
=(2+√2)/(1+√2+1/2)
=2(2+√2)/(3+2√2)
=2√2(√2+1)/(√2+1)²
=2√2/(√2+1)
=2√2(√2-1)
=4-2√2