If {x^2} + {y^2} + {z^2} = {r^2},then {tan ^{ | vedclass

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(B) Given the expression: $S = {	an ^{ - 1}}\left( {\frac{{xy}}{{zr}}} ight) + {	an ^{ - 1}}\left( {\frac{{yz}}{{xr}}} ight) + {	an ^{ - 1}}\le

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$\frac{\pi }{2}$

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(B) Given the expression: $S = {	an ^{ - 1}}\left( {\frac{{xy}}{{zr}}} ight) + {	an ^{ - 1}}\left( {\frac{{yz}}{{xr}}} ight) + {	an ^{ - 1}}\left( {\frac{{zx}}{{yr}}} ight)$.Using the identity ${	an ^{ - 1}}A + {	an ^{ - 1}}B + {	an ^{ - 1}}C = {	an ^{ - 1}}\left( {\frac{{A + B + C - ABC}}{{1 - (AB + BC + CA)}}} ight)$,we have:$A = \frac{{xy}}{{zr}}, B = \frac{{yz}}{{xr}}, C = \frac{{zx}}{{yr}}$.Then $A + B + C = \frac{{x^2y^2 + y^2z^2 + z^2x^2}}{{xyzr}}$ and $ABC = \frac{{x^2y^2z^2}}{{xyzr^3}} = \frac{{xyz}}{{r^3}}$.Also,$AB + BC + CA = \frac{{xy^2z}}{{xr^2}} + \frac{{yz^2x}}{{yr^2}} + \frac{{zx^2y}}{{zr^2}} = \frac{{y^2 + z^2 + x^2}}{{r^2}} = \frac{{r^2}}{{r^2}} = 1$.Since the denominator $1 - (AB + BC + CA) = 1 - 1 = 0$,the argument of ${	an ^{ - 1}}$ approaches $\infty$.Therefore,$S = {	an ^{ - 1}}(\infty) = \frac{\pi }{2}$.

If ${x^2} + {y^2} + {z^2} = {r^2}$,then ${\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{zx}}{{yr}}} \right) = $

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