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If $\mathrm{e}^{\mathrm{x}}=\mathrm{y}+\sqrt{1+\mathrm{y}^{2}}$, then the value of $\mathrm{y}$ is
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Verified Answer
The correct answer is:
$\frac{1}{2}\left(\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}\right)$
Given $\mathrm{e}^{\mathrm{x}}=\mathrm{y}+\sqrt{1+\mathrm{y}^{2}}$ $\Rightarrow \mathrm{e}^{\mathrm{x}}-\mathrm{y}=\sqrt{1+\mathrm{y}^{2}}$
Squaring both side, we have
$\begin{array}{l}
\mathrm{e}^{2 \mathrm{x}}+\mathrm{y}^{2}-2 \mathrm{e}^{\mathrm{x}} \mathrm{y}=1+\mathrm{y}^{2} \\
\Rightarrow 2 \mathrm{e}^{\mathrm{x}} \mathrm{y}=\mathrm{e}^{2 \mathrm{x}}-1 \\
\Rightarrow \mathrm{y}=\frac{\mathrm{e}^{2 \mathrm{x}}-1}{2 \mathrm{e}^{\mathrm{x}}} \Rightarrow \mathrm{y}=\frac{1}{2}\left[\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}\right]
\end{array}$
Squaring both side, we have
$\begin{array}{l}
\mathrm{e}^{2 \mathrm{x}}+\mathrm{y}^{2}-2 \mathrm{e}^{\mathrm{x}} \mathrm{y}=1+\mathrm{y}^{2} \\
\Rightarrow 2 \mathrm{e}^{\mathrm{x}} \mathrm{y}=\mathrm{e}^{2 \mathrm{x}}-1 \\
\Rightarrow \mathrm{y}=\frac{\mathrm{e}^{2 \mathrm{x}}-1}{2 \mathrm{e}^{\mathrm{x}}} \Rightarrow \mathrm{y}=\frac{1}{2}\left[\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}\right]
\end{array}$
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