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Question Number 13626 by tawa tawa last updated on 21/May/17

$$\mathrm{Prove}\mathrm{that}$$$${\mathrm{cosech}}^{-1}\left(\mathrm{x}\right)={\sinh }^{-1}\left(\frac{1}{\mathrm{x}}\right)$$

Answered by mrW1 last updated on 21/May/17

$$u={\mathrm{cosech}}^{-1}\left(\mathrm{x}\right)$$$$\Rightarrow x=\frac{2}{e^u-e^{-u}}$$$$\Rightarrow \frac{1}{x}=\frac{e^u-e^{-u}}{2}$$$$v={\sinh }^{-1}\left(\frac{1}{\mathrm{x}}\right)$$$$\Rightarrow \frac{1}{x}=\frac{e^v-e^{-v}}{2}$$$$$$$$\Rightarrow \frac{e^u-e^{-u}}{2}=\frac{e^v-e^{-v}}{2}$$$$\Rightarrow u=v$$$$\Rightarrow {\mathrm{cosech}}^{-1}\left(\mathrm{x}\right)={\sinh }^{-1}\left(\frac{1}{\mathrm{x}}\right)$$

Commented by tawa tawa last updated on 21/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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