根號下1+tan^2α=√(1/cosα)^2=|1/cosα|
同理根號下1+cot^2α=|1/sinα|
原式=cosα|cosα|+sinα|sinα|=-1
而(cosα)^2+(sinα)^2
所以|cosα|=-cosα,|sinα|=-sinα
也就是cosα,sinα都是負數(shù),正余弦函數(shù)都是負值,則α在第三象限
cosα|cosα|+sinα|sinα|可能是
(cosα)^2+(sinα)^2=1
-(cosα)^2-(sinα)^2=-1
(cosα)^2-(sinα)^2=cos2α無窮多個值
-(cosα)^2+(sinα)^2=-cos2α還是無窮多個值