sinh^{-1}left(frac{x}{sqrt{1-x^2}}right) is e | vedclass

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(D) Let $\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight) = 	heta$. Then,$\sinh 	heta = \frac{x}{\sqrt{1-x^2}}$. We know the identity $\cosh^2 	heta -

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$	anh^{-1} x$

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(D) Let $\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight) = 	heta$. Then,$\sinh 	heta = \frac{x}{\sqrt{1-x^2}}$. We know the identity $\cosh^2 	heta - \sinh^2 	heta = 1$,so $\cosh^2 	heta = 1 + \sinh^2 	heta$. Substituting the value of $\sinh 	heta$: $\cosh^2 	heta = 1 + \left(\frac{x}{\sqrt{1-x^2}}ight)^2 = 1 + \frac{x^2}{1-x^2} = \frac{1-x^2+x^2}{1-x^2} = \frac{1}{1-x^2}$. Thus,$\cosh 	heta = \frac{1}{\sqrt{1-x^2}}$. Now,$	anh 	heta = \frac{\sinh 	heta}{\cosh 	heta} = \frac{x/\sqrt{1-x^2}}{1/\sqrt{1-x^2}} = x$. Therefore,$	heta = 	anh^{-1} x$. Hence,$\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight) = 	anh^{-1} x$.

$\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ is equal to

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