If {e^x} = y + sqrt {1 + {y^2}} ,then y = | vedclass

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(B) Given that ${e^x} = y + \sqrt{1 + y^2}$.Rearranging the terms,we get ${e^x} - y = \sqrt{1 + y^2}$.Squaring both sides,we obtain $({e^x} - y)^2 = 1

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$\frac{{{e^x} - {e^{ - x}}}}{2}$

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(B) Given that ${e^x} = y + \sqrt{1 + y^2}$.Rearranging the terms,we get ${e^x} - y = \sqrt{1 + y^2}$.Squaring both sides,we obtain $({e^x} - y)^2 = 1 + y^2$.Expanding the left side,${e^{2x}} + y^2 - 2y{e^x} = 1 + y^2$.Subtracting $y^2$ from both sides,${e^{2x}} - 2y{e^x} = 1$.Rearranging to solve for $y$,$2y{e^x} = {e^{2x}} - 1$.Dividing by $2{e^x}$,$y = \frac{{e^{2x}} - 1}{2{e^x}}$.Simplifying the expression,$y = \frac{1}{2} \left( \frac{e^{2x}}{e^x} - \frac{1}{e^x} \right) = \frac{e^x - e^{-x}}{2}$.

If ${e^x} = y + \sqrt {1 + {y^2}} $,then $y =$

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