Question

If ${\tan }^{-1}x+{\tan }^{-1}+{\tan }^{-1}z=\frac{\pi }{2}$, then

• A. $x+y+z-xyz=0$
• B. $x+y+z+xyz=0$
• C. $xy+yz+zx+1=0$
• D. $xy+yz+zx-1=0$

Answer 1

The correct option is D $xy+yz+zx-1=0$

${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)=\frac{\pi }{2}-{\tan }^{-1}\left(z\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{x+y}{1-xy}\right)={\cot }^{-1}\left(z\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{x+y}{1-xy}\right)={\tan }^{-1}\left(\frac{1}{z}\right)$

Taking $\tan$ on both the sides, we get

$z(x+y)=1-xy$

$\Rightarrow xz+yz=1-xy$

$\Rightarrow xy+yz+zx-1=0$