If xy>1 then prove that tan^-1x+tan^-1y=π+tan^-1(x+y)÷(1-xy)

(i) tan^-1x + tan^-1y = tan^-1((x+y)/(1-xy)), xy .

Let tan^-1x = A and tan^-1y = B

Þ tan A = x and tan B = y

Now, tan(A+B) = (tanA + tanB)/(1-tanAtanB)

⇒ tan(A+B) = (x+y)/(1-xy)

⇒tan^-1[(x+y)/(1-xy)] = A + B

⇒tan^-1[(x+y)/(1-xy)] = tan^-1x + tan^-1y

(ii) tan^-1x - tan^-1y = tan^-1((x-y)/(1+xy)), xy > -1.

Proof:

Let tan^-1x = α ⇒ x = tan α

     tan^-1y = β ⇒ y = tan β

Now, tan(α - β) = (tan α - tan β)/(1+ tanαtanβ) = (x-y)/(1+xy)

α - β = tan^-1 [(x-y)/(1+xy)]

tan^-1x - tan^-1y = tan^-1 [(x-y)/(1+xy)]

Hence, tan^-1x - tan^-1y = tan^-1 [(x-y)/(1+xy)], xy > -1