Value of x Satisfying Inverse Trigonometric Equation

To find the value of $x \in (0, 1)$ satisfying the given equation $\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1 - x^2})$

we simplify both sides independently.

Step 1: Simplify the Left-Hand Side (LHS) Let $\theta = \tan^{-1}(x\sqrt{2})$. Since $x \in (0, 1)$, $\theta \in \left(0, \frac{\pi}{2}\right)$ and $\tan \theta = x\sqrt{2}$.

Using the trigonometric identity $\sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}$, we get: $\text{LHS} = \sin(\theta) = \frac{x\sqrt{2}}{\sqrt{1 + (x\sqrt{2})^2}} = \frac{x\sqrt{2}}{\sqrt{1 + 2x^2}}$

Step 2: Simplify the Right-Hand Side (RHS) Let $\phi = \sin^{-1}\sqrt{1 - x^2}$. Since $x \in (0, 1)$, $\sqrt{1 - x^2} \in (0, 1)$, so $\phi \in \left(0, \frac{\pi}{2}\right)$ and $\sin \phi = \sqrt{1 - x^2}$.

The cosine of $\phi$ is given by: $\cos \phi = \sqrt{1 - \sin^2 \phi} = \sqrt{1 - (1 - x^2)} = \sqrt{x^2} = x \quad (\text{since } x > 0)$

Thus, the cotangent of $\phi$ is: $\text{RHS} = \cot \phi = \frac{\cos \phi}{\sin \phi} = \frac{x}{\sqrt{1 - x^2}}$

Step 3: Solve the Equation Equating LHS and RHS: $\frac{x\sqrt{2}}{\sqrt{1 + 2x^2}} = \frac{x}{\sqrt{1 - x^2}}$

Since $x \in (0, 1)$, $x \neq 0$. Dividing both sides by $x$: $\frac{\sqrt{2}}{\sqrt{1 + 2x^2}} = \frac{1}{\sqrt{1 - x^2}}$

Squaring both sides (as all quantities are positive): $\frac{2}{1 + 2x^2} = \frac{1}{1 - x^2}$

Cross-multiplying gives: $2(1 - x^2) = 1 + 2x^2$ $2 - 2x^2 = 1 + 2x^2$ $4x^2 = 1$ $x^2 = \frac{1}{4}$

Since $x \in (0, 1)$, taking the positive square root yields: $x = \frac{1}{2}$

Thus, the value of $x$ is $\frac{1}{2}$.

Correct Option: A