关于任意角的三角函数的一些性质问题?

tanx+cotx
=sinx/cosx +cosx/sinx
=[(sinx)^2+(cosx)^2]/sinxcosx
=1/(sinxcosx)
1+tan^2(x)
=1+sin^2(x)/cos^2(x)
=cos^2(x)/cos^2(x)+sin^2(x)/cos^2(x)
=[sin^2(x)+cos^2(x)]/cos^2(x)
=1/cos^2(x)
1+cot^2(x)
=1+cos^2(x)/sin^2(x)
=sin^2(x)/sin^2(x)+cos^2(x)/sin^2(x)
=[sin^2(x)+cos^2(x)]/sin^2(x)
=1/sin^2(x)
tanxcotx
=[sinx/cosx]*[cosx/sinx]
=1

tanX=sinX/cosX
sin{-X}=-sinX
cos{-X}=cosX
tan{-X]=-tanX
sin2X=2sinXcosX
cos2X=2cox2X-1
tan2X=sin2Xcos2X
......