જો {x^2} + {y^2} + {z^2} = {r^2} હોય,તો {tan | vedclass

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(B) આપેલ પદાવલિ: $S = {	an ^{ - 1}}\left( {\frac{{xy}}{{zr}}} ight) + {	an ^{ - 1}}\left( {\frac{{yz}}{{xr}}} ight) + {	an ^{ - 1}}\left( {\fra

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$\frac{\pi }{2}$

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(B) આપેલ પદાવલિ: $S = {	an ^{ - 1}}\left( {\frac{{xy}}{{zr}}} ight) + {	an ^{ - 1}}\left( {\frac{{yz}}{{xr}}} ight) + {	an ^{ - 1}}\left( {\frac{{zx}}{{yr}}} ight)$.નિત્યસમ ${	an ^{ - 1}}A + {	an ^{ - 1}}B + {	an ^{ - 1}}C = {	an ^{ - 1}}\left( {\frac{{A + B + C - ABC}}{{1 - (AB + BC + CA)}}} ight)$ નો ઉપયોગ કરતા:$A = \frac{{xy}}{{zr}}, B = \frac{{yz}}{{xr}}, C = \frac{{zx}}{{yr}}$.અહીં $AB + BC + CA = \frac{{xy^2z}}{{xr^2}} + \frac{{yz^2x}}{{yr^2}} + \frac{{zx^2y}}{{zr^2}} = \frac{{y^2 + z^2 + x^2}}{{r^2}} = \frac{{r^2}}{{r^2}} = 1$.છેદ $1 - (AB + BC + CA) = 1 - 1 = 0$ હોવાથી,${	an ^{ - 1}}$ નો આર્ગ્યુમેન્ટ $\infty$ તરફ જાય છે.તેથી,$S = {	an ^{ - 1}}(\infty) = \frac{\pi }{2}$.

જો ${x^2} + {y^2} + {z^2} = {r^2}$ હોય,તો ${\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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