If $sin\left( an^{- 1}(x\sqrt{2}) ight) = cot\left( \sin^{- 1}\sqrt{1 - x^{2}} ight),x \in (0,

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

Accepted Answer

Sol. [Key Concept: Convert both inverse-trig sides to algebraic ratios: $\frac{x\sqrt{2}}{\sqrt{2x^{2} + 1}} = \frac{1}{\sqrt{1 - x^{2}}}$, square and solve for $x$.]$${\frac{x\sqrt{2}}{\sqrt{2x^{2} + 1}} = \frac{1}{\sqrt{1 - x^{2}}} }{2\left( 1 - x^{2} \right) = \left( 2x^{2} + 1 \right) }{2 - 2x^{2} = 2x^{2} + 1 }{x^{2} = \frac{1}{4} }{x = \frac{1}{2} }$$

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