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

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

Grade:11

2 Answers

Umar
12 Points
4 years ago
There is an error in this question. By mistake I had left π in the right hand side. I had also posted the right question 
Arun
25750 Points
4 years ago

(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

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