Question Number 43180 by gunawan last updated on 08/Sep/18
$$\mathrm{The}\:\mathrm{smallest}\:\mathrm{and}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{values}\:\mathrm{of} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\:,\:\:\mathrm{0}\leqslant\:{x}\:\leqslant\:\mathrm{1}\:\:\mathrm{are} \\ $$
Commented by maxmathsup by imad last updated on 08/Sep/18
![let ϕ(x)=arctan(((1−x)/(1+x))) with x∈[0,1] changement x =cosθ give ϕ(x)=ϕ(cosθ) =arctan(((1−cosθ))/(1+cosθ))) =arctan( tan^2 ((θ/2))) we have 0≤x≤1 ⇒0≤θ≤(π/2) ⇒ 0≤tan((π/2))≤1 ⇒0≤tan^2 ((θ/2))≤1 ⇒ 0≤ϕ(cosθ)≤(π/4) ⇒ ϕ(x)∈[0,(π/4)](both function tan and arctan are increasing)](Q43242.png)
$${let}\:\varphi\left({x}\right)={arctan}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\:\:\:{with}\:{x}\in\left[\mathrm{0},\mathrm{1}\right]\:{changement}\:{x}\:={cos}\theta\:{give} \\ $$$$\varphi\left({x}\right)=\varphi\left({cos}\theta\right)\:={arctan}\left(\frac{\left.\mathrm{1}−{cos}\theta\right)}{\mathrm{1}+{cos}\theta}\right)\:={arctan}\left(\:{tan}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\right) \\ $$$${we}\:{have}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\mathrm{0}\leqslant{tan}\left(\frac{\pi}{\mathrm{2}}\right)\leqslant\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant{tan}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\leqslant\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{0}\leqslant\varphi\left({cos}\theta\right)\leqslant\frac{\pi}{\mathrm{4}}\:\Rightarrow\:\varphi\left({x}\right)\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]\left({both}\:{function}\:{tan}\:{and}\:{arctan}\:{are}\:{increasing}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18
$${y}={tan}^{−\mathrm{1}} \mathrm{1}−{tan}^{−\mathrm{1}} {x} \\ $$$${y}=\frac{\Pi}{\mathrm{4}}−{tan}^{−\mathrm{1}} {x} \\ $$$${when}\:{x}=\mathrm{0} \\ $$$${tan}^{−\mathrm{1}} {x}=\mathrm{0} \\ $$$${y}=\frac{\Pi}{\mathrm{4}} \\ $$$${when}\:{x}=\mathrm{1}\:\:\: \\ $$$${tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\frac{\Pi}{\mathrm{4}} \\ $$$${so}\:{y}=\frac{\Pi}{\mathrm{4}}−\frac{\Pi}{\mathrm{4}}=\mathrm{0} \\ $$$${so}\:{max}\:{of}\:{y}=\frac{\Pi}{\mathrm{4}}\:\:\:\:{min}\:{of}\:{y}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by gunawan last updated on 08/Sep/18
$$\mathrm{Nice}\:\mathrm{Sir} \\ $$