If the value of x satisfying the equation sin | vedclass

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(B) Let $\sin^{-1} \sqrt{1-x^2} = 	heta$. Then $\sin 	heta = \sqrt{1-x^2}$.This implies $\cos 	heta = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = |x|$. Since

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$5$

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(B) Let $\sin^{-1} \sqrt{1-x^2} = 	heta$. Then $\sin 	heta = \sqrt{1-x^2}$.This implies $\cos 	heta = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = |x|$. Since the domain of $\sqrt{\frac{2}{x}-1}$ requires $x > 0$,we have $\cos 	heta = x$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sqrt{1-x^2}}{x} = \sqrt{\frac{1-x^2}{x^2}} = \sqrt{\frac{1}{x^2}-1}$.Given $	an 	heta = \sqrt{\frac{2}{x}-1}$,we equate the two expressions:$\sqrt{\frac{1}{x^2}-1} = \sqrt{\frac{2}{x}-1}$.Squaring both sides: $\frac{1}{x^2} - 1 = \frac{2}{x} - 1$.$\frac{1}{x^2} = \frac{2}{x} \implies x^2 = \frac{x}{2} \implies x(x - \frac{1}{2}) = 0$.Since $x > 0$,we have $x = \frac{1}{2}$.Here,$a = 1$ and $b = 2$,which are coprime.Therefore,$a^2 + b^2 = 1^2 + 2^2 = 1 + 4 = 5$.

If the value of $x$ satisfying the equation $\sin^{-1} \sqrt{1-x^2} = \tan^{-1} \sqrt{\frac{2}{x}-1}$ is $\frac{a}{b}$ (where $a$ and $b$ are coprime),then the value of $a^2 + b^2$ is

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