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if in triangle ABC a=6,b=3 and cos(A-B)=4/5 then its area is
Dear Hemant cos (A-B) = (1 -tan²(A-B)/2)/(1+tan²(A-B)/2) = 4/5On cross multiplication and then solving tan²(A-B)/2 = 1/9tan(A-B)/2 = 1/3 Hence(a-b) cot(C/2)/(a+b) = 1/3On solving Cot(c/2) = 1C/2 = 45°C = 90°Hence it is a right angle triangle Now area = (1/2)*3*6 = 9 square units RegardsArun (askIITians forum expert)
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