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

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Inverse Trigonometric Functions

sin (tan^{- 1} (x 2)) = cot (sin^{- 1} 1 - x^{2}), x \in (0, 1)

, then the value of x is :

Answer: (A)

With

θ = tan^{- 1} (x 2)

tan θ = x 2

, so

sin θ = \frac{x 2}{1 + 2 x ^{2}}

. With

ϕ = sin^{- 1} 1 - x^{2}

sin ϕ = 1 - x^{2}

and

cos ϕ = 1 - (1 - x^{2}) = x

for

x \in (0, 1)

, so

cot ϕ = \frac{x}{1 - x ^{2}}

. The equation becomes

\frac{x 2}{1 + 2 x ^{2}} = \frac{x}{1 - x ^{2}}

. Cancelling

x

and cross-multiplying,

2 1 - x^{2} = 1 + 2 x^{2}

. Squaring,

2 (1 - x^{2}) = 1 + 2 x^{2}

, hence

4 x^{2} = 1

and

x = \frac{1}{2}

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