If $x + y + z = x y z$ , and $x, y, z > 0$ , then find the value of ${tan}^{- 1} x + {tan}^{- 1} y + {tan}^{- 1} z$

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Solution

Let $x = t a n A$ , $y = t a n B$ and $z = t a n C$ such that $A + B + C = π$ .
Then
$x + y + z$
$= t a n A + t a n B + t a n C$
$= (t a n A + t a n B) + t a n C$
$= t a n (A + B) (1 - t a n A . t a n B) + t a n C$
$= t a n (π - C) (1 - t a n A . t a n B) + t a n C$
$= t a n C - t a n C (1 - t a n A . t a n B)$
$= t a n C (1 - 1 + t a n A . t a n B)$
$= t a n A . t a n B . t a n C$
$= x y z$ .
Hence
$t a n^{- 1} (x) + t a n^{- 1} (y) + t a n^{- 1} (z)$
$= t a n^{- 1} (t a n A) + t a n^{- 1} (t a n B) + t a n^{- 1} (t a n C)$
$= A + B + C$
$= π$ ... (as taken).

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