Question

If ${\tan }^{-1}x+{\tan }^{1}y+{\tan }^{1}z=\frac{\pi }{2}$, then prove that, $xy+yz+zx=1$.

Answer 1

${\tan }^{1}x+{\tan }^{1}y+{\tan }^{1}z=\frac{\pi }{2}$
By taking L.H.S.
$={\tan }^{1}x+{\tan }^{1}y+{\tan }^{1}z$
$={\tan }^1\left(\frac{x+y}{1-xy}\right)+{\tan }^{1}z$
$[\because {\tan }^{1}x+{\tan }^{1}y={\tan }^1\left(\frac{x+y}{1-xy}\right)]$
$={\tan }^1\left(\frac{\frac{x+y}{1-xy}+z}{1-\frac{(x+y)}{1-xy}.z}\right)$
$[\because {\tan }^{1}x+{\tan }^{1}y={\tan }^1\left(\frac{x+y}{1-xy}\right)]$
$={\tan }^1(\frac{x+y+z-xyz}{1-xy-xz-yz}\times \frac{1-xy}{1-xy})$
$={\tan }^1\left(\frac{x+y+z-xyz}{1-xy-xz-yz}\right)$
$\because {\tan }^{1}x+{\tan }^{1}y+{\tan }^{1}z=\frac{\pi }{2}$
$\therefore {\tan }^1\left(\frac{x+y+z-xyz}{1-xy-xz-yz}\right)=\frac{\pi }{2}$
$\frac{x+y+z-xyz}{1-xy-xz-yz}=\tan \frac{\pi }{2}$
$\frac{x+y+z-xyz}{1-xy-xz-yz}=\frac{1}{0}[\because \tan \frac{\pi }{2}=\infty =\frac{1}{0}]$
$0\times [x+y+z-xyz]=[1-xy-yz-xz]\times 1$
$0=1-xy-yz-xz$
$xy+yz+zx=1$.