A+b=?, if tan(a)=3, tan(b)= -1/2 , 0

Question

A+b=?, if tan(a)=3, tan(b)= -1/2 , 0

Answer 1

Найдем tan(a+b) = (tan(a)+tan(b))/(1-tan(a)*tan(b))
1) tan(a+b) = 2.5/(1+1.5) = 1. Значит a+b = pi/4 + pi*n
2) tan(a+b) = 3/(1-0) = 3. Значит a+b = arctan(3)+pi*n