A+B+C=90°,证明tanAtanB+tanBtanC+tanCtanA=1（详细过程）

tanAtanB+tanBtanC+tanCtanA=tanAtanB+(tanA+tanB)ctan(B+A) (by A+B+C=90°)
=(sinAsinB)/(cosAcosB)+[(sinA/cosA)+(sinB/cosB)]cos(A+B)/sin(A+B)
=(sinAsinB)/(cosAcosB)+[(sinAcosB+sinB+cosA)/(cosAcosB)]cos(A+B)/sin(A+B)
=(sinAsinB)/(cosAcosB)+cos(A+B)/(cosAcosB)
=1