If sin(tan^{-1}(xsqrt{2})) = cot(sin^{-1}sqrt | vedclass

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(A) Let $	an^{-1}(x\sqrt{2}) = 	heta$. Then $	an 	heta = x\sqrt{2}$.Using the identity $\sin 	heta = \frac{	an 	heta}{\sqrt{1+	an^2 	heta}}$,

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$1/2$

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(A) Let $	an^{-1}(x\sqrt{2}) = 	heta$. Then $	an 	heta = x\sqrt{2}$.Using the identity $\sin 	heta = \frac{	an 	heta}{\sqrt{1+	an^2 	heta}}$, we get $\sin 	heta = \frac{x\sqrt{2}}{\sqrt{1+2x^2}}$.Let $\sin^{-1}(\sqrt{1-x^2}) = \phi$. Then $\sin \phi = \sqrt{1-x^2}$.Using the identity $\cot \phi = \frac{\cos \phi}{\sin \phi} = \frac{\sqrt{1-\sin^2 \phi}}{\sin \phi}$, we get $\cot \phi = \frac{\sqrt{1-(1-x^2)}}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}}$.Equating the two sides: $\frac{x\sqrt{2}}{\sqrt{1+2x^2}} = \frac{x}{\sqrt{1-x^2}}$.Since $x \in (0, 1)$, $x 
eq 0$, we can divide by $x$: $\frac{\sqrt{2}}{\sqrt{1+2x^2}} = \frac{1}{\sqrt{1-x^2}}$.Squaring both sides: $\frac{2}{1+2x^2} = \frac{1}{1-x^2}$.$2(1-x^2) = 1+2x^2 \implies 2-2x^2 = 1+2x^2 \implies 4x^2 = 1 \implies x^2 = 1/4$.Since $x \in (0, 1)$, $x = 1/2$.

If $\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1-x^2})$ for $x \in (0, 1)$, then the value of $x$ is:

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