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$x=\log \left(\frac{1}{y}+\sqrt{1+\frac{1}{y^2}}\right) \Rightarrow y$ is equal to
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Verified Answer
The correct answer is:
$\operatorname{cosech} x$
Given,
$x=\log \left(\frac{1}{y}+\sqrt{1+\frac{1}{y^2}}\right)$
$\begin{array}{ll}\therefore & x=\operatorname{cosech}^{-1} y \\ \Rightarrow & y=\operatorname{cosech} x\end{array}$
$x=\log \left(\frac{1}{y}+\sqrt{1+\frac{1}{y^2}}\right)$
$\begin{array}{ll}\therefore & x=\operatorname{cosech}^{-1} y \\ \Rightarrow & y=\operatorname{cosech} x\end{array}$
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