Problem 3 - Entrance Test

If tan(A) + tan(B) = a and cot(A) + cot(B) = b, then cot(A+B) =

A. ab/(a+b)B. b/(ab)C. a/(ab)D. (a+b)/ab

Correct: A

Given tan(A) + tan(B) = a and cot(A) + cot(B) = b => 1/tan(A) + 1/tan(B) = b => (tan(A) + tan(B)) / (tan(A)tan(B)) = b => a / (tan(A)tan(B)) = b => tan(A)tan(B) = a/b. Now, cot(A+B) = (1 - tan(A)tan(B)) / (tan(A) + tan(B)) = (1 - a/b) / a = (b-a) / (ab) = (b/ab)-(a/ab). From given equation cot(A)+cot(B) = b => (tanA+tanB)/tanAtanB = b=> a/(tanAtanB) = b => tanAtanB= a/b. Hence cot(A+B) = (1-tanAtanB)/(tanA+tanB) = (1-(a/b))/a = (b-a)/ab