tan(a+b)=(tana+tanb)/(1-tanatanb),即tana+tanb=tan(a+b)(1-tanatanb),
所以tanatanb+tanbtany+tanytana
=tanatanb+tany(tana+tanb)
=tanatanb+(tana+tanb)/tan(a+b)
=tanatanb+tan(a+b)(1-tanatanb)/tan(a+b)
=tanatanb+(1-tanatanb)
=1