MCQ
If $\tan ^{ - 1}x + \tan ^{ - 1}y + \tan ^{ - 1}z = \frac{\pi }{2},$ then
  • A
    $x + y + z - xyz = 0$
  • B
    $x + y + z + xyz = 0$
  • C
    $xy + yz + zx + 1 = 0$
  • $xy + yz + zx - 1 = 0$

Answer

Correct option: D.
$xy + yz + zx - 1 = 0$
d
(d) Given that $\tan ^{ - 1}x + \tan ^{ - 1}y + \tan ^{ - 1}z = \frac{\pi }{2},$

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left[ {\frac{{x + y + z - xyz}}{{1 - xy - yz - xz}}} \right] = \frac{\pi }{2}$

$ \Rightarrow \,\,\left[ {\frac{{x + y + z - xyz}}{{1 - xy - yz - zx}}} \right] = \tan \frac{\pi }{2} = \frac{1}{0}$

Hence $xy + yz + zx - 1 = 0$.

Trick : $x = y = z = \frac{1}{{\sqrt 3 }},$ so that

${\tan ^{ - 1}}\frac{1}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{1}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{1}{{\sqrt 3 }} = \frac{\pi }{2}$

Obviously $ (d)$ holds for these values of  $x, y, z.$

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