If 2 sinh^{-1}left(frac{a}{sqrt{1-a^2}}right) | vedclass

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

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$a$

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(A) We know that $\sinh^{-1}(y) = \log(y + \sqrt{1+y^2})$.Let $y = \frac{a}{\sqrt{1-a^2}}$. Then $\sqrt{1+y^2} = \sqrt{1 + \frac{a^2}{1-a^2}} = \sqrt{\frac{1-a^2+a^2}{1-a^2}} = \frac{1}{\sqrt{1-a^2}}$.So,$\sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}ight) = \log\left(\frac{a}{\sqrt{1-a^2}} + \frac{1}{\sqrt{1-a^2}}ight) = \log\left(\frac{a+1}{\sqrt{1-a^2}}ight) = \log\left(\frac{1+a}{\sqrt{(1-a)(1+a)}}ight) = \log\left(\sqrt{\frac{1+a}{1-a}}ight) = \frac{1}{2} \log\left(\frac{1+a}{1-a}ight)$.Given equation is $2 \sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}ight) = \log \left(\frac{1+x}{1-x}ight)$.Substituting the value,we get $2 	imes \frac{1}{2} \log\left(\frac{1+a}{1-a}ight) = \log \left(\frac{1+x}{1-x}ight)$.$\log\left(\frac{1+a}{1-a}ight) = \log \left(\frac{1+x}{1-x}ight)$.Comparing both sides,we get $x = a$.

If $2 \sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$,then $x$ is equal to

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