The value of x which satisfies the equation t | vedclass

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(A) Given the equation: $	an^{-1}x = \sin^{-1}\left(\frac{3}{\sqrt{10}}ight)$.Let $	heta = \sin^{-1}\left(\frac{3}{\sqrt{10}}ight)$. Then $\sin\

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$3$

Vedclass · Answer

(A) Given the equation: $	an^{-1}x = \sin^{-1}\left(\frac{3}{\sqrt{10}}ight)$.Let $	heta = \sin^{-1}\left(\frac{3}{\sqrt{10}}ight)$. Then $\sin	heta = \frac{3}{\sqrt{10}}$.Using the identity $\cos^2	heta = 1 - \sin^2	heta$,we get $\cos^2	heta = 1 - \left(\frac{3}{\sqrt{10}}ight)^2 = 1 - \frac{9}{10} = \frac{1}{10}$.Thus,$\cos	heta = \frac{1}{\sqrt{10}}$.Now,$	an	heta = \frac{\sin	heta}{\cos	heta} = \frac{3/\sqrt{10}}{1/\sqrt{10}} = 3$.Therefore,$	heta = 	an^{-1}(3)$.Substituting this back into the original equation: $	an^{-1}x = 	an^{-1}(3)$.Hence,$x = 3$.

The value of $x$ which satisfies the equation $\tan^{-1}x = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right)$ is

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