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Question Number 13626 by tawa tawa last updated on 21/May/17
Prove that  cosech^(−1) (x) = sinh^(−1) ((1/x))
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{cosech}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$
Answered by mrW1 last updated on 21/May/17
u=cosech^(−1) (x)   ⇒x=(2/(e^u −e^(−u) ))  ⇒(1/x)=((e^u −e^(−u) )/2)  v= sinh^(−1) ((1/x))  ⇒(1/x)=((e^v −e^(−v) )/2)    ⇒((e^u −e^(−u) )/2)=((e^v −e^(−v) )/2)  ⇒u=v  ⇒cosech^(−1) (x) = sinh^(−1) ((1/x))
$${u}=\mathrm{cosech}^{−\mathrm{1}} \left(\mathrm{x}\right)\: \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}}{{e}^{{u}} −{e}^{−{u}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{{e}^{{u}} −{e}^{−{u}} }{\mathrm{2}} \\ $$$${v}=\:\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{{e}^{{v}} −{e}^{−{v}} }{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\frac{{e}^{{u}} −{e}^{−{u}} }{\mathrm{2}}=\frac{{e}^{{v}} −{e}^{−{v}} }{\mathrm{2}} \\ $$$$\Rightarrow{u}={v} \\ $$$$\Rightarrow\mathrm{cosech}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$
Commented by tawa tawa last updated on 21/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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