# Prove that (tan A + tan B)^2 = [((tan(A +B))^2 + (tan A)^2(tan B)^2(tan(A +B))^2 - 2((tan A)(tan B)(tan(A +B))^2)] We have to prove (tan A + tan B)^2 = [((tan(A +B))^2 + (tan A)^2(tan B)^2(tan(A +B))^2 - 2((tan A)(tan B)(tan(A +B))^2)]

An easy way to do this is to start with tan (A + B) = (tan A + tan B)/(1 - (tan A)(tan B))

=> (1 - (tan...

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We have to prove (tan A + tan B)^2 = [((tan(A +B))^2 + (tan A)^2(tan B)^2(tan(A +B))^2 - 2((tan A)(tan B)(tan(A +B))^2)]

An easy way to do this is to start with tan (A + B) = (tan A + tan B)/(1 - (tan A)(tan B))

=> (1 - (tan A)(tan B))* tan(A +B) = tan A + tan B

=> tan A + tan B = (1 - (tan A)(tan B))* tan(A +B)

square both the sides

=> (tan A + tan B)^2 = [(1 - (tan A)(tan B))]^2* [tan(A +B)]^2

=> (tan A + tan B)^2 = [(1 + (tan A)^2(tan B)^2 - 2((tan A)(tan B))] * [tan(A +B)]^2

=> (tan A + tan B)^2 = [((tan(A +B))^2 + (tan A)^2(tan B)^2(tan(A +B))^2 - 2((tan A)(tan B)(tan(A +B))^2)]

Therefore (tan A + tan B)^2 = [((tan(A +B))^2 + (tan A)^2(tan B)^2(tan(A +B))^2 - 2((tan A)(tan B)(tan(A +B))^2)]

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