sir i have a question: If in triangle ABC, (tanA-tanB)/(tanA+tanB) = (c-b)/c, prove that A= 60 degree.
(tanA-tanB?/(tanA+tanB) = (c-b)/c
sin(A-B)/sinC =(sinC-sinB)/sinC (Since, sin(A+B)=sinC)
sinB = sin(A+B)-sin(A-B) = 2sinBcosA
Thus, angle A = 60 degrees
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