Question

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

Answer 1

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

Let ${\tan }^{-1}x=\alpha$

$\Rightarrow \tan \alpha =x$

${\tan }^{-1}y=\beta$

$\Rightarrow \tan \beta =y$

${\tan }^{-1}z=\gamma$

$\Rightarrow \tan \gamma =z$

$\Rightarrow \alpha +\beta +\gamma =\frac{\pi }{2}$

$\tan (\alpha +\beta +\gamma )=\frac{\tan \alpha +\tan \beta +\tan \gamma -\tan \alpha \tan \beta \tan \gamma }{1-(\tan \alpha \tan \beta +\tan \beta \tan \gamma +\tan \gamma \tan \alpha )}=\tan \frac{\pi }{2}$

$=$ not defined

$\Rightarrow$ Denominator$=0$

$\Rightarrow \tan \alpha \tan \beta +\tan \beta \tan \gamma +\tan \gamma \tan \alpha =1$

$\Rightarrow xy+yz+zx=1$.