Question Number 177811 by Spillover last updated on 09/Oct/22
![](https://www.tinkutara.com/question/27255.png)
Answered by CElcedricjunior last updated on 09/Oct/22
![Tan x=sinhπ=((e^π βe^(βπ) )/2) =>e^(2π) β2e^π tanxβ1=0 π«=4tan^2 x+4=(4/(cos^2 x)) =>e^π =((2tanxβ(2/(cosx)))/2)=tanxβ(1/(cosx)) =>π=ln(tanxβsecx) si xβ](π/2);((3π)/2)[ =>π=ln(tanx+secx)si xβ]β (π/2);(π/2)[ ........le celebre cedric junior...............](https://www.tinkutara.com/question/Q177830.png)
$$\boldsymbol{\mathrm{Tan}}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{sinh}\theta}=\frac{\boldsymbol{{e}}^{\boldsymbol{\theta}} β\boldsymbol{{e}}^{β\boldsymbol{\theta}} }{\mathrm{2}} \\ $$$$=>\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\theta}} β\mathrm{2}\boldsymbol{{e}}^{\boldsymbol{\theta}} \boldsymbol{\mathrm{tanx}}β\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\Delta}=\mathrm{4}\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{{x}}+\mathrm{4}=\frac{\mathrm{4}}{\boldsymbol{\mathrm{co}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{s}}{x}}} \\ $$$$=>\boldsymbol{{e}}^{\boldsymbol{\theta}} =\frac{\mathrm{2}\boldsymbol{\mathrm{tan}{x}}\mp\frac{\mathrm{2}}{\boldsymbol{{cosx}}}}{\mathrm{2}}=\boldsymbol{{tanx}}\mp\frac{\mathrm{1}}{{cos}\boldsymbol{{x}}} \\ $$$$\left.=>\boldsymbol{\theta}=\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{tanx}}β\boldsymbol{{secx}}\right)\:\boldsymbol{\mathrm{si}}\:\boldsymbol{\mathrm{x}}\in\right]\frac{\boldsymbol{\pi}}{\mathrm{2}};\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2}}\left[\right. \\ $$$$\left.=>\boldsymbol{\theta}=\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{tanx}}+\boldsymbol{{secx}}\right)\boldsymbol{\mathrm{si}}\:\boldsymbol{\mathrm{x}}\in\right]β\:\frac{\boldsymbol{\pi}}{\mathrm{2}};\frac{\boldsymbol{\pi}}{\mathrm{2}}\left[\right. \\ $$$$\: \\ $$$$……..{le}\:{celebre}\:{cedric}\:{junior}…………… \\ $$$$ \\ $$