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Question Number 177811 by Spillover last updated on 09/Oct/22

Answered by CElcedricjunior last updated on 09/Oct/22

Tan x=sinh𝛉=((e^𝛉 βˆ’e^(βˆ’π›‰) )/2) =>e^(2𝛉) βˆ’2e^𝛉 tanxβˆ’1=0 𝚫=4tan^2 x+4=(4/(cos^2 x)) =>e^𝛉 =((2tanxβˆ“(2/(cosx)))/2)=tanxβˆ“(1/(cosx)) =>𝛉=ln(tanxβˆ’secx) si x∈](𝛑/2);((3𝛑)/2)[ =>𝛉=ln(tanx+secx)si x∈]βˆ’ (𝛑/2);(𝛑/2)[ ........le celebre cedric junior...............

$$\boldsymbol{\mathrm{Tan}}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{sinh}\theta}=\frac{\boldsymbol{{e}}^{\boldsymbol{\theta}} βˆ’\boldsymbol{{e}}^{βˆ’\boldsymbol{\theta}} }{\mathrm{2}} \\ $$$$=>\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\theta}} βˆ’\mathrm{2}\boldsymbol{{e}}^{\boldsymbol{\theta}} \boldsymbol{\mathrm{tanx}}βˆ’\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\Delta}=\mathrm{4}\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{{x}}+\mathrm{4}=\frac{\mathrm{4}}{\boldsymbol{\mathrm{co}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{s}}{x}}} \\ $$$$=>\boldsymbol{{e}}^{\boldsymbol{\theta}} =\frac{\mathrm{2}\boldsymbol{\mathrm{tan}{x}}\mp\frac{\mathrm{2}}{\boldsymbol{{cosx}}}}{\mathrm{2}}=\boldsymbol{{tanx}}\mp\frac{\mathrm{1}}{{cos}\boldsymbol{{x}}} \\ $$$$\left.=>\boldsymbol{\theta}=\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{tanx}}βˆ’\boldsymbol{{secx}}\right)\:\boldsymbol{\mathrm{si}}\:\boldsymbol{\mathrm{x}}\in\right]\frac{\boldsymbol{\pi}}{\mathrm{2}};\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2}}\left[\right. \\ $$$$\left.=>\boldsymbol{\theta}=\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{tanx}}+\boldsymbol{{secx}}\right)\boldsymbol{\mathrm{si}}\:\boldsymbol{\mathrm{x}}\in\right]βˆ’\:\frac{\boldsymbol{\pi}}{\mathrm{2}};\frac{\boldsymbol{\pi}}{\mathrm{2}}\left[\right. \\ $$$$\: \\ $$$$........{le}\:{celebre}\:{cedric}\:{junior}............... \\ $$$$ \\ $$

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