Question

If xy + yz + zx = 1, then :

• A. ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=0$
• B. ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\pi$
• C. ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\frac{\pi }{4}$
• D. ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\frac{\pi }{2}$

Answer 1

Answer: (D)

$xy+yz+zx=1$ ......(1)
Now, we know ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z$
$={\tan }^{-1}\left[\frac{x+y+z-xyz}{1-\left(xy+yz+zx\right)}\right]$
using equation (1) we have
${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z$
$={\tan }^{-1}\left(\frac{1}{0}\right)=\tan \infty$
$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\frac{\pi }{2}$