Question

Let $x^{2}+y^{2}+z^{2}=r^{2}.$ If $tan^{- 1}\left(\frac{x y}{z r}\right)+tan^{- 1}\left(\frac{y z}{x r}\right)+tan^{- 1}\left(\frac{z x}{y r}\right)=m\pi $ then evaluate $2m$ .

Answer 1

$tan^{- 1}\left(\frac{x y}{z r}\right)+tan^{- 1}\left(\frac{y z}{x r}\right)+tan^{- 1}\left(\frac{z x}{y r}\right)$
$=tan^{- 1}\left[\frac{\frac{x y}{2 r} + \frac{y z}{x r} + \frac{z x}{y r} - \frac{x y z}{r^{3}}}{1 - \left(\frac{x^{2} + y^{2} + z^{2}}{r^{2}}\right)}\right]$
$x^{2}+y^{2}+z^{2}=r^{2}$
$\Rightarrow tan^{- 1}\left(\frac{x y}{z r}\right)+tan^{- 1}\left(\frac{y z}{x r}\right)+tan^{- 1}\left(\frac{z x}{y r}\right)=\frac{\pi }{2}$
$\Rightarrow m=\frac{1}{2}=0.5\Rightarrow 2m=1$