prove that tan^(−1) (((xy)/(rz)))+tan^(−1) (((xz)/(ry)))+tan^(−1) (((yz)/(rx)))=(π/2)


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Question Number 157932 by cortano last updated on 30/Oct/21

$p r o v e t h a t \tan^{- 1} (\frac{x y}{r z}) + \tan^{- 1} (\frac{x z}{r y}) + \tan^{- 1} (\frac{y z}{r x}) = \frac{π}{2}$
Commented by cortano last updated on 30/Oct/21

$n o s i r . o n l y i t c o n d i t i o n$
Commented by som(math1967) last updated on 30/Oct/21

$A n y o t h e r c o n d i t i o n ?$
Commented by som(math1967) last updated on 30/Oct/21
$let tan^(−1) (((xy)/(rz)))+tan^(−1) (((yz)/(rx)))+tan^(−1) (((zx)/(ry)))=(π/2) tan^(−1) (((((xy)/(rz))+((yz)/(rx)))/(1−((xy)/(rz)).((yz)/(rx)))))=(π/2) −tan^(−1) (((zx)/(ry))) tan^(−1) (((x^2 y+yz^2 )/(rzx))×((r^2 zx)/(r^2 zx−xy^2 z)))=(π/2)−tan^(−1) (((zx)/(ry))) ((y(x^2 +z^2 )r)/(zx(r^2 −y^2 )))=tan{(π/2)−tan^(−1) (((zx)/(ry)))} ((yr(x^2 +z^2 ))/(zx(r^2 −y^2 )))=cot{tan^(−1) (((zx)/(ry)))} ((yr(x^2 +z^2 ))/(zx(r^2 −y^2 )))=cot{cot^(−1) (((ry)/(zx)))} ((yr(x^2 +z^2 ))/(zx(r^2 −y^2 )))=((ry)/(zx)) x^2 +z^2 =r^2 −y^2 ∴x^2 +y^2 +z^2 =r^2 so I think x^2 +y^2 +z^2 =r^2 is the condition$
$l e t \tan^{- 1} (\frac{x y}{r z}) + \tan^{- 1} (\frac{y z}{r x}) + \tan^{- 1} (\frac{z x}{r y}) = \frac{π}{2}$ $\tan^{- 1} (\frac{\frac{x y}{r z} + \frac{y z}{r x}}{1 - \frac{x y}{r z} . \frac{y z}{r x}}) = \frac{π}{2} - \tan^{- 1} (\frac{z x}{r y})$ $\tan^{- 1} (\frac{x^{2} y + {y z}^{2}}{r z x} \times \frac{r^{2} z x}{r^{2} z x - {x y}^{2} z}) = \frac{π}{2} - \tan^{- 1} (\frac{z x}{r y})$ $\frac{y (x^{2} + z^{2}) r}{z x (r^{2} - y^{2})} = t a n {\frac{π}{2} - \tan^{- 1} (\frac{z x}{r y})}$ $\frac{y r (x^{2} + z^{2})}{z x (r^{2} - y^{2})} = c o t {\tan^{- 1} (\frac{z x}{r y})}$ $\frac{y r (x^{2} + z^{2})}{z x (r^{2} - y^{2})} = c o t {\cot^{- 1} (\frac{r y}{z x})}$ $\frac{y r (x^{2} + z^{2})}{z x (r^{2} - y^{2})} = \frac{r y}{z x}$ $x^{2} + z^{2} = r^{2} - y^{2}$ $∴ x^{2} + y^{2} + z^{2} = r^{2}$ $s o I t h i n k x^{2} + y^{2} + z^{2} = r^{2} i s t h e c o n d i t i o n$
Commented by cortano last updated on 30/Oct/21

$t h a n k s s i r$
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