in a triangle ABC, tanA + tanB + tanC=6 and tanAtanB=2, then find the values of tanA,tanB and tanC
Hello Matin, given tan A + tan B + tan C = 6
But C = 180 -(A+B)
So tan C = tan {180 -(A+B)} = - tan (A+B) --(1)
Now tan (A +B) = tan A + tan B / (1 - tan A tan B) = - (tan A + tan B) since tan A tan B = 2 (given) --(2)
So (1) becomes tan C = tan A + tan B
Plug in first and you have 2 (tan A + tan B) = 6
Or tan A + tan B = 3 --(3)
Now tan C = 3
Again from (2) and (3) we have tan A = 1 and tan B = 2
But C = 180 -(A+B)
So tan C = tan {180 -(A+B)} = - tan (A+B) --(1)
Now tan (A +B) = tan A + tan B / (1 - tan A tan B) = - (tan A + tan B) since tan A tan B = 2 (given) --(2)
So (1) becomes tan C = tan A + tan B
Plug in first and you have 2 (tan A + tan B) = 6
Or tan A + tan B = 3 --(3)
Now tan C = 3
Again from (2) and (3) we have tan A = 1 and tan B = 2
