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If $x=\log \left(y+\sqrt{y^2+1}\right)$ then $y=$
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The correct answer is:
$\sin h \mathrm{x}$
$x=\log \left(y+\sqrt{\left.y^2+1\right)}\right.$
$\begin{aligned} & y+\sqrt{y^2+1}=e^x \\ & \because \sin h^{-1} x=\log \left(x+\sqrt{x^2+1}\right) \\ & \sin h^{-1} y=\log \left(y+\sqrt{y^2+1}\right) \\ & \sin h^{-1} y=x \Rightarrow \sin h x=y\end{aligned}$
$\begin{aligned} & y+\sqrt{y^2+1}=e^x \\ & \because \sin h^{-1} x=\log \left(x+\sqrt{x^2+1}\right) \\ & \sin h^{-1} y=\log \left(y+\sqrt{y^2+1}\right) \\ & \sin h^{-1} y=x \Rightarrow \sin h x=y\end{aligned}$
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