If a

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(D) We know that $\operatorname{Sinh}^{-1}(y) = \log(y + \sqrt{y^2+1})$.Let $y = \frac{a}{\sqrt{1-a^2}}$.Then $y^2+1 = \frac{a^2}{1-a^2} + 1 = \frac{a

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$a$

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(D) We know that $\operatorname{Sinh}^{-1}(y) = \log(y + \sqrt{y^2+1})$.Let $y = \frac{a}{\sqrt{1-a^2}}$.Then $y^2+1 = \frac{a^2}{1-a^2} + 1 = \frac{a^2+1-a^2}{1-a^2} = \frac{1}{1-a^2}$.So,$\sqrt{y^2+1} = \frac{1}{\sqrt{1-a^2}}$ (since $a Thus,$\operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}ight) = \log\left(\frac{a+1}{\sqrt{1-a^2}}ight) = \log\left(\frac{a+1}{\sqrt{(1-a)(1+a)}}ight) = \log\left(\sqrt{\frac{1+a}{1-a}}ight) = \frac{1}{2} \log\left(\frac{1+a}{1-a}ight)$.Given equation: $2 \operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}ight) = \log \left(\frac{1+x}{1-x}ight)$.Substituting the derived value: $2 	imes \frac{1}{2} \log\left(\frac{1+a}{1-a}ight) = \log \left(\frac{1+x}{1-x}ight)$.Therefore,$\log\left(\frac{1+a}{1-a}ight) = \log \left(\frac{1+x}{1-x}ight)$.Comparing both sides,we get $x = a$.

If $a < 1$ and $2 \operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$,then $x=$

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