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Question
Prove that `tan^-1x + tan^-1y + tan^-1z = tan^-1[(x + y + z - xyz)/(1 - xy - yz - zx)]`
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Solution
`tan^-1x + tan^-1y = tan^-1 ((x + y)/(1 - xy))`
= `tan^-1 ("A")`
Here A = `(x + y)/(1 - xy)`
So L.H.S: `tan^-1x + tan^-1y + tan^-1z = tan^-1 ("A") + tan^-1z`
`tan^-1 (("A" + z)/(1 - "A"z)) = tan^-1 [((x + y)/(1 - xy + z))/(1 - (x + y)/(1 - xy) (z))]`
= `tan^-1 [((x + y + z(1 - xy))/(1 - xy))/((1 - xy - (x + y)z)/(1 - xy))]`
= `tan^-1 [(x + y + z - xyz)/(1 - xy - xz - yz)]`
= R.H.S
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