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Question Number 30939 by paddu1234 last updated on 01/Mar/18

If x^2 +y^2 +z^2 = r^2 , then tan^(−1) (((xy)/(zr)))+tan^(−1) (((yz)/(xr)))+tan^(−1) (((xz)/(yr))) =

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

Answered by ajfour last updated on 01/Mar/18

let tan θ = ((xy)/(zr)) , tan φ = ((yz)/(xr)) , and tan ψ = ((xz)/(yr)) ⇒ tan (θ+φ+ψ)=((((xy)/(zr))+((yz)/(xr))+((zx)/(yr))−((xyz)/r^3 ))/(1−((y^2 /r^2 )+(z^2 /r^2 )+(x^2 /r^2 )))) ⇒ tan (θ+φ+ψ)= not defined hence θ+φ+ψ=±(π/2) , or may be = ±((3π)/2) .

$${let}\:\:\mathrm{tan}\:\theta\:=\:\frac{{xy}}{{zr}}\:\:,\:\mathrm{tan}\:\phi\:=\:\frac{{yz}}{{xr}}\:, \\ $$$${and}\:\:\:\mathrm{tan}\:\psi\:=\:\frac{{xz}}{{yr}} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\left(\theta+\phi+\psi\right)=\frac{\frac{{xy}}{{zr}}+\frac{{yz}}{{xr}}+\frac{{zx}}{{yr}}−\frac{{xyz}}{{r}^{\mathrm{3}} }}{\mathrm{1}−\left(\frac{{y}^{\mathrm{2}} }{{r}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{{r}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\:\mathrm{tan}\:\left(\theta+\phi+\psi\right)=\:{not}\:{defined} \\ $$$${hence}\:\:\:\theta+\phi+\psi=\pm\frac{\pi}{\mathrm{2}}\:,\:{or}\:{may}\:{be} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\pm\frac{\mathrm{3}\pi}{\mathrm{2}}\:\:\:. \\ $$

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