Prove:sech1 x = cosh1(1/x), 0 < x 1coth1 x = tanh1(1/x),

Chapter 6, Problem 62

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Prove:

$\operatorname{sech}^{-1} x=\cosh ^{-1}(1 / x), \quad 0<x \leq 1$

$\operatorname{coth}^{-1} x=\tanh ^{-1}(1 / x), \quad|x|>1$

$\operatorname{csch}^{-1} x=\sinh ^{-1}(1 / x), \quad x \neq 0$

Equation Transcription:

${sech}^{-1}\ x\ ={cosh}^{-1}(1/x),\ \ \ \ \ \ \ 0\ <\ x\ \leq{}1$

${coth}^{-1}\ x\ ={tanh}^{-1}(1/x),\ \ \ \ \ \ \ |x|>1$

${csch}^{-1}\ x\ ={sinh}^{-1}(1/x),\ \ \ \ \ \ \ \ x\ \ne{}\ 0\$

Text Transcription:

sech^-1 x=cosh^-1 (1/x), 0 < x leq 1

coth^-1 x=tanh^-1 (1/x), |x|>1

csch^1 x=sinh^-1 (1/x), x neq 0

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