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

Answered by CElcedricjunior last updated on 09/Oct/22

$$\mathbf{Tan}\mathbf{x}=\mathbf{sinh}\mathit{\theta }=\frac{{\mathit{e}}^{\mathit{\theta }}-{\mathit{e}}^{-\mathit{\theta }}}{2}$$$$=>{\mathbf{e}}^{2\mathit{\theta }}-2{\mathit{e}}^{\mathit{\theta }}\mathbf{tanx}-1=0$$$$\mathbf{\Delta }=4{\mathbf{tan}}^2\mathit{x}+4=\frac{4}{\mathbf{co}\stackrel{2}{\mathbf{s}x}}$$$$=>{\mathit{e}}^{\mathit{\theta }}=\frac{2\mathbf{tan}\mathit{x}\mp \frac{2}{\mathit{c}\mathit{o}\mathit{s}\mathit{x}}}{2}=\mathit{t}\mathit{a}\mathit{n}\mathit{x}\mp \frac{1}{cos\mathit{x}}$$$$=>\mathit{\theta }=\mathbf{ln}(\mathbf{tanx}-\mathit{s}\mathit{e}\mathit{c}\mathit{x})\mathbf{si}\mathbf{x}\in ]\frac{\mathit{\pi }}{2};\frac{3\mathit{\pi }}{2}[$$$$=>\mathit{\theta }=\mathbf{ln}(\mathbf{tanx}+\mathit{s}\mathit{e}\mathit{c}\mathit{x})\mathbf{si}\mathbf{x}\in ]-\frac{\mathit{\pi }}{2};\frac{\mathit{\pi }}{2}[$$$$$$$$........lecelebrecedricjunior...............$$$$$$

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