（1）已知sinx+cosx=根号2/2,求sin^4x+cos^4x

sinx+cosx=√2/2
(sinx+cosx)^2=1/2
(sinx)^2+2sinxcosx+(cosx)^2=1/2
1+2sinxcosx=1/2
2sinxcosx=-1/2
sinxcosx=-1/4
sin^4x+cos^4x
=sin^4x+2(sinxcosx)^2+cos^4x -2(sinxcosx)^2
=[sin^2x+cos^2x ]^2-2(sinxcosx)^2
=1^2-2*(1/4)^2
=1-1/8
=7/8
√[(1+tanx)^2+（1-tanx)^2]
=√[1+tan^2x+2tanx+1+tan^2x-2tanx]
=√[2+2tan^2x]
=√[2(1+tan^2x)]
=√[2sec^2x)]
=-secx√2
2cosθ/√(1-sin^2θ）+√(1-cos^2θ)/sinθ
=2√[cos^2θ/(1-sin^2θ]+√[(1-cos^2θ)/sin^2θ]
=2√[cos^2θ/cos^2θ]+√[sin^2θ/sin^2θ]
=2+1
=3