Question Number 157932 by cortano last updated on 30/Oct/21
$${prove}\:{that}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{xy}}{{rz}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{xz}}{{ry}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{yz}}{{rx}}\right)=\frac{\pi}{\mathrm{2}} \\ $$
Commented by cortano last updated on 30/Oct/21
$${no}\:{sir}.\:{only}\:{it}\:{condition} \\ $$
Commented by som(math1967) last updated on 30/Oct/21
$${Any}\:{other}\:{condition}\:? \\ $$
Commented by som(math1967) last updated on 30/Oct/21
$${let}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{xy}}{{rz}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{yz}}{{rx}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{{xy}}{{rz}}+\frac{{yz}}{{rx}}}{\mathrm{1}−\frac{{xy}}{{rz}}.\frac{{yz}}{{rx}}}\right)=\frac{\pi}{\mathrm{2}}\:−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} {y}+{yz}^{\mathrm{2}} }{{rzx}}×\frac{{r}^{\mathrm{2}} {zx}}{{r}^{\mathrm{2}} {zx}−{xy}^{\mathrm{2}} {z}}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right) \\ $$$$\frac{{y}\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right){r}}{{zx}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}={tan}\left\{\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right)\right\} \\ $$$$\frac{{yr}\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{{zx}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}={cot}\left\{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right)\right\} \\ $$$$\frac{{yr}\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{{zx}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}={cot}\left\{\mathrm{cot}^{−\mathrm{1}} \left(\frac{{ry}}{{zx}}\right)\right\} \\ $$$$\frac{{yr}\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{{zx}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}=\frac{{ry}}{{zx}} \\ $$$${x}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\therefore{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${so}\:{I}\:{think}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} \:{is}\:{the}\:{condition} \\ $$
Commented by cortano last updated on 30/Oct/21
$${thanks}\:{sir} \\ $$