Menu Close

If-xy-yz-zx-1-then-tan-1-x-tan-1-y-tan-1-z-

Tinku Tara 0 Comments
Pin




Question Number 48527 by Pk1167156@gmail.com last updated on 25/Nov/18
If xy + yz + zx = 1, then tan^(−1) x + tan^(−1) y + tan^(−1) z =
$$\mathrm{If}\:\:{xy}\:+\:{yz}\:+\:{zx}\:=\:\mathrm{1},\:\mathrm{then} \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}\:+\:\mathrm{tan}^{−\mathrm{1}} {y}\:+\:\mathrm{tan}^{−\mathrm{1}} {z}\:=\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18
tan^(−1) x+tan^(−1) (((y+z)/(1−yz))) =tan^(−1) (((x+((y+z)/(1−yz)))/(1−x×((y+z)/(1−yz))))) =tan^(−1) (((x−xyz+y+z)/(1−yz−xy−xz))) tan^(−1) (((x+y+z−xyz)/0)) =tan^(−1) (∞) =tan^(−1) (tan(π/2)) =(π/2)
$${tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} \left(\frac{{y}+{z}}{\mathrm{1}−{yz}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{{y}+{z}}{\mathrm{1}−{yz}}}{\mathrm{1}−{x}×\frac{{y}+{z}}{\mathrm{1}−{yz}}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{{x}−{xyz}+{y}+{z}}{\mathrm{1}−{yz}−{xy}−{xz}}\right) \\ $$$${tan}^{−\mathrm{1}} \left(\frac{{x}+{y}+{z}−{xyz}}{\mathrm{0}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\infty\right) \\ $$$$={tan}^{−\mathrm{1}} \left({tan}\frac{\pi}{\mathrm{2}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}} \\ $$
Answered by ajfour last updated on 25/Nov/18
tan^(−1) x = θ , tan^(−1) y = φ , tan^(−1) z = ψ let δ = θ+φ+ψ tan δ = tan (θ+φ^(−) +ψ) = ((tan (θ+φ)+tan ψ)/(1−tan (θ+φ)tan ψ)) = ((((tan θ+tan φ)/(1−tan θtan φ))+tan ψ)/(1−tan (θ+φ)tan ψ)) = ((Σtan θ−Πtan θ)/(1−Σtan θ tan φ)) but Σtan θ tan φ = Σxy = 1 ⇒ 𝛅 = θ+φ+ψ = ±(π/2) .
$$\mathrm{tan}^{−\mathrm{1}} {x}\:=\:\theta\:,\:\mathrm{tan}^{−\mathrm{1}} {y}\:=\:\phi\:,\:\mathrm{tan}^{−\mathrm{1}} {z}\:=\:\psi \\ $$$${let}\:\:\:\delta\:=\:\theta+\phi+\psi \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\delta\:=\:\mathrm{tan}\:\left(\overline {\theta+\phi}+\psi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{tan}\:\left(\theta+\phi\right)+\mathrm{tan}\:\psi}{\mathrm{1}−\mathrm{tan}\:\left(\theta+\phi\right)\mathrm{tan}\:\psi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\frac{\mathrm{tan}\:\theta+\mathrm{tan}\:\phi}{\mathrm{1}−\mathrm{tan}\:\theta\mathrm{tan}\:\phi}+\mathrm{tan}\:\psi}{\mathrm{1}−\mathrm{tan}\:\left(\theta+\phi\right)\mathrm{tan}\:\psi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\Sigma\mathrm{tan}\:\theta−\Pi\mathrm{tan}\:\theta}{\mathrm{1}−\Sigma\mathrm{tan}\:\theta\:\mathrm{tan}\:\phi} \\ $$$${but}\:\:\Sigma\mathrm{tan}\:\theta\:\mathrm{tan}\:\phi\:=\:\Sigma{xy}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:\boldsymbol{\delta}\:=\:\theta+\phi+\psi\:=\:\pm\frac{\pi}{\mathrm{2}}\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *