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Question: Answered & Verified by Expert
$x=\log \left(\frac{1}{y}+\sqrt{1+\frac{1}{y^2}}\right) \Rightarrow y$ is equal to
MathematicsInverse Trigonometric FunctionsAP EAMCETAP EAMCET 2012
Options:
  • A $\tanh x$
  • B $\operatorname{coth} x$
  • C $\operatorname{sech} x$
  • D $\operatorname{cosech} x$
Solution:
2465 Upvotes Verified Answer
The correct answer is: $\operatorname{cosech} x$
Given,
$x=\log \left(\frac{1}{y}+\sqrt{1+\frac{1}{y^2}}\right)$
$\begin{array}{ll}\therefore & x=\operatorname{cosech}^{-1} y \\ \Rightarrow & y=\operatorname{cosech} x\end{array}$

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