在锐角三角形中,求证三个内角正切之和等于三个内角正切之积

假设三个内角分别是A,B,C,则C=180-A-B
所以tanC = tan((90-A) + (90-B)) = [tan(90-A) + tan(90-B)] / [1 - tan(90-A)tan(90-B)]
= (ctgA + ctgB) / (1-ctgActgB) 《－－上下同时乘以tanAtanB
= (tanA + tanB) / (tanAtanB - 1)
所以tanA + tanB + tanC
= tanA + tanB + (tanA + tanB) / (tanAtanB - 1)
= (tanA + tanB) * [1 + 1 / (tanAtanB - 1)]
= (tanA + tanB) * [tanAtanB / (tanAtanB - 1)]
= tanA * tanB * (tanA + tanB) / (tanAtanB - 1)
= tanA * tanB * tanC

A,B,C,
tanA+tanB+tanC=tan(A+B)*(1-tanAtanB)+tanC=
=tan(A+B)-tan(A+B)tanAtanB+tanC tan(A+B)=-tanC
=-tanC+tanCtanBtanC+tanC
=tanCtanBtanC
三个内角正切之和等于三个内角正切之积