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Question
If tan−1x + tan−1y + tan−1z = π, then show that `1/(xy) + 1/(yz) + 1/(zx)` = 1
Solution
tan−1x + tan−1y + tan−1z = π
∴ tan−1x + tan−1y = π − tan−1z
∴ `tan^-1 ((x + y)/(1 - xy))` = π − tan−1z
∴ `(x + y)/(1 - xy)` = tan(π − tan−1z)
∴ `(x + y)/(1 - xy)` = −tan(tan−1z)
∴ `(x + y)/(1 - xy)` = − z
∴ x + y = −z + xyz
∴ x + y + z = xyz
∴ `1/(yz) + 1/(xz) + 1/(xy)` = 1, i.e., `1/(xy) + 1/(yz) + 1/(zx)` = 1
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