Proof for inverse trigonometric formula

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I am denoting tan inverse of x as tan^-1(x) and tan inverse of y as tan^-1(y) and tan inverse of (x-y)/(1+xy) as tan^-1((x-y)/(1+xy)).

Required to prove :

tan^-1((x-y)/(1+xy)) = tan^-1(x) - tan^-1(y), for x>0 and y>0

Proof :

Let x>0 and y>0 and

let tan^-1(x) = A and tan^-1(y) = B -------(1)

Here, A belongs to (0,π/2) and B belongs to (0,π/2). [Since x>0 and y>0]

And tan(A) = x and tan(B) = y [From(1)]

Here, (A-B) belongs to (-π/2,π/2).

Reason for (A-B) belongs to (-π/2,π/2) :

Let (A-B) is greater than or equal to π/2

Then A will be greater than or equal to (π/2)+B. Then A should be greater than π/2 but A belongs to (0,π/2).

Therefore, 

(A-B) is less than π/2

In similar way, if we do, we get that (A-B) is greater than -π/2.


Consider 

tan(A-B) = (tan(A)-tan(B))/(1+tan(A)tan(B))

Therefore, tan(A-B) = (x-y)/(1+xy) [From above]

As tan^-1((x-y)/(1+xy)) belongs to (-π/2,π/2) and (A-B)  belongs to (-π/2,π/2),

tan^-1((x-y)/(1+xy)) = A-B

Therefore,

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

Therefore,

tan^-1((x-y)/(1+xy)) = tan^-1(x) - tan^-1(y), for x>0 and y>0





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