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

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

$${prove}\:{that}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{xy}}{{rz}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{xz}}{{ry}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{yz}}{{rx}}\right)=\frac{\pi}{\mathrm{2}} \\ $$

Commented by cortano last updated on 30/Oct/21

no sir. only it condition

$${no}\:{sir}.\:{only}\:{it}\:{condition} \\ $$

Commented by som(math1967) last updated on 30/Oct/21

Any other condition ?

$${Any}\:{other}\:{condition}\:? \\ $$

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

$${let}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{xy}}{{rz}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{yz}}{{rx}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{{xy}}{{rz}}+\frac{{yz}}{{rx}}}{\mathrm{1}−\frac{{xy}}{{rz}}.\frac{{yz}}{{rx}}}\right)=\frac{\pi}{\mathrm{2}}\:−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} {y}+{yz}^{\mathrm{2}} }{{rzx}}×\frac{{r}^{\mathrm{2}} {zx}}{{r}^{\mathrm{2}} {zx}−{xy}^{\mathrm{2}} {z}}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right) \\ $$$$\frac{{y}\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right){r}}{{zx}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}={tan}\left\{\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right)\right\} \\ $$$$\frac{{yr}\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{{zx}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}={cot}\left\{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{zx}}{{ry}}\right)\right\} \\ $$$$\frac{{yr}\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{{zx}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}={cot}\left\{\mathrm{cot}^{−\mathrm{1}} \left(\frac{{ry}}{{zx}}\right)\right\} \\ $$$$\frac{{yr}\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{{zx}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}=\frac{{ry}}{{zx}} \\ $$$${x}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\therefore{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${so}\:{I}\:{think}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} \:{is}\:{the}\:{condition} \\ $$

Commented by cortano last updated on 30/Oct/21

thanks sir

$${thanks}\:{sir} \\ $$

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