假設三個內角分別是A,B,C，則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) 《－－上下同時乘以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