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Question
If `tan^-1 "x" + tan^-1 "y" + tan^-1 "z" = pi/2,` then show that xy + yz + zx = 1
Solution
`tan^-1 "x" + tan^-1 "y" + tan^-1 "z" = pi/2`
∴ `tan^-1 (("x + y")/(1 - "xy")) + tan^-1 "z" = pi/2`
∴ `tan^-1 [(("x + y")/(1 - "xy") + "z")/(1 - (("x + y")/(1 - "xy"))"z")] = pi/2`
∴ `tan^-1 [("x + y + z - xyz")/(1 - xy - xz - yz)] = pi/2`
∴ `("x + y + z - xyz")/(1 - "xy" - "yz" - "zx") = tan pi/2`, which does not exist
∴ 1 - xy - yz - zx = 0
∴ xy + yz + zx = 1
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