求A+B+C的值?
tan(A+B)
=(tanA+tanB)/(1-tanA*tanB)
=(1/2+1/5)/(1-1/2*1/5)
=7/9
tan(A+B+C)
=(tan(A+B)+tanC)/(1-tan(A+B)*tanC)
=(7/9+1/8)/(1-7/9*1/8)
=1
因A、B、C都是锐角,且因为正切值