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

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$	anh^{-1} x$

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(D) Let $	heta = \sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight)$.Then,$\sinh 	heta = \frac{x}{\sqrt{1-x^2}}$.We know that $\cosh^2 	heta = 1 + \sinh^2 	heta$.Substituting the value of $\sinh 	heta$:$\cosh^2 	heta = 1 + \left(\frac{x}{\sqrt{1-x^2}}ight)^2 = 1 + \frac{x^2}{1-x^2} = \frac{1-x^2+x^2}{1-x^2} = \frac{1}{1-x^2}$.Therefore,$\cosh 	heta = \frac{1}{\sqrt{1-x^2}}$ (taking the positive root as $\cosh 	heta > 0$).Now,$	anh 	heta = \frac{\sinh 	heta}{\cosh 	heta} = \frac{x/\sqrt{1-x^2}}{1/\sqrt{1-x^2}} = x$.Thus,$	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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