BV1rY4y1x78F

有

cosA/(1+sinA)

=sin(π/2-A)/(1+cos(π/2-A)

=tan(π/4-A/2)

且

sin2B/(1+cos2B)

=tanB

即

tan(π/4-A/2)=tanB

即

A/2+B=π/4

即

A=π/2-2B

C=B+π/2

即

0＜π/2-2B＜π

且

0＜B＜π

且

0＜B+π/2＜π

即

-π/4＜B＜π/4

且

0＜B＜π

且

-π/2＜B＜π/2

即

0＜B＜π/4

設(shè)

cos2B=t

t∈(1/2，1）

有

原式

=(sin2A+sin2B）/sin2C

=(sin2(π/2-2B）+sin2B）/sin2(B+π/2）

=(cos22B+sin2B)/cos2B

=((2t-1)2-t+1）/t

=(4t2-5t+2)/t

=4t+2/t-5

≥4√2-5