tan ^{-1}left(frac{1}{2 sqrt{2}}right)+sin ^{ | vedclass

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(A) Given: $	an ^{-1}\left(\frac{1}{2 \sqrt{2}}ight)+\sin ^{-1}\left(\frac{1}{\sqrt{3}}ight)=\cos ^{-1} x$.Let $	heta_1 = 	an ^{-1} \left(\frac

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$\frac{1}{\sqrt{3}}$

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(A) Given: $	an ^{-1}\left(\frac{1}{2 \sqrt{2}}ight)+\sin ^{-1}\left(\frac{1}{\sqrt{3}}ight)=\cos ^{-1} x$.Let $	heta_1 = 	an ^{-1} \left(\frac{1}{2 \sqrt{2}}ight)$ and $	heta_2 = \sin ^{-1} \left(\frac{1}{\sqrt{3}}ight)$.From the definition of inverse trigonometric functions:$	an 	heta_1 = \frac{1}{2 \sqrt{2}} \implies \cos 	heta_1 = \frac{2 \sqrt{2}}{3}$ and $\sin 	heta_1 = \frac{1}{3}$.$\sin 	heta_2 = \frac{1}{\sqrt{3}} \implies \cos 	heta_2 = \sqrt{1 - \left(\frac{1}{\sqrt{3}}ight)^2} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.Now, $	heta_1 + 	heta_2 = \cos ^{-1} x \implies x = \cos(	heta_1 + 	heta_2)$.Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:$x = \cos 	heta_1 \cos 	heta_2 - \sin 	heta_1 \sin 	heta_2$$x = \left(\frac{2 \sqrt{2}}{3}ight) \left(\frac{\sqrt{2}}{\sqrt{3}}ight) - \left(\frac{1}{3}ight) \left(\frac{1}{\sqrt{3}}ight)$$x = \frac{2 	imes 2}{3 \sqrt{3}} - \frac{1}{3 \sqrt{3}} = \frac{4}{3 \sqrt{3}} - \frac{1}{3 \sqrt{3}} = \frac{3}{3 \sqrt{3}} = \frac{1}{\sqrt{3}}$.Thus, $x = \frac{1}{\sqrt{3}}$.

$\tan ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)+\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\cos ^{-1} x$, then $x=$

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