Question

If $r^2=x^2+y^2+z^2$, then prove that
${\tan }^{-1}\left(\frac{yz}{rx}\right)+{\tan }^{-1}\left(\frac{zx}{ry}\right)+{\tan }^{-1}\left(\frac{xy}{rz}\right)=\frac{\pi }{2}$

Answer 1

Given $r^2=x^2+y^2+z^2.......\left(i\right)$
LHS
$={\tan }^{-1}\left(\frac{yz}{rx}\right)+{\tan }^{-1}\left(\frac{zx}{ry}\right)+{\tan }^{-1}\left(\frac{xy}{rz}\right)$
$={\tan }^{-1}\left(\frac{S_1-S_3}{{1-S}_2}\right)$

$S_1,S_2$ and $S_3$ are the sum of tan x , tan y , tan z taken 1, 2 and 3 at a time respectively.

Now,
$1-S_2=1-(\frac{yz}{rx}\times \frac{zx}{ry}+\frac{zx}{ry}\times \frac{xy}{rz}+\frac{xy}{rz}\times \frac{yz}{rx})$
$=1-(\frac{z^2}{r^2}+\frac{x^2}{r^2}+\frac{y^2}{r^2})$
$=1-\frac{\left(r^2\right)}{r^2}=1-1=0$
$\therefore \theta ={\tan }^{-1}\left(\frac{S_1-S_3}{0}\right)\Rightarrow \theta =\frac{\pi }{2}$