If sin(tan^-1(xsqrt(2))) = cot(sin^-1(sqrt(1-x^2))), x is in (0, 1), then the value of x is:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2026

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Home Maths Inverse TrigonometryQuestion #230

\sin (\tan^{- 1} (x \sqrt{2})) = \cot (\sin^{- 1} \sqrt{1 - x^{2}})

x \in (0, 1)

, then the value of

x

is:

Clue 1 of 3: Inverse Trigonometric Identities

Simplify each side of the equation by converting the inverse trigonometric functions into standard trigonometric ratios.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Simplify the Left Hand Side (LHS)✦ Active

Let $θ = \tan^{- 1} (x \sqrt{2})$ . Then $\tan θ = x \sqrt{2}$ . Construct a right triangle with opposite side $x \sqrt{2}$ and adjacent side $1$ . The hypotenuse is $\sqrt{(x \sqrt{2})^{2} + 1^{2}} = \sqrt{2 x^{2} + 1}$ .

\sin (\tan^{- 1} (x \sqrt{2})) = \sin θ = \frac{x \sqrt{2}}{\sqrt{2 x^{2} + 1}}

Step 2: Simplify the Right Hand Side (RHS)○ Expand

Let $ϕ = \sin^{- 1} \sqrt{1 - x^{2}}$ . Then $\sin ϕ = \sqrt{1 - x^{2}}$ . Since $x \in (0, 1)$ , $ϕ \in (0, \frac{π}{2})$ . Using the identity $\cos^{2} ϕ = 1 - \sin^{2} ϕ$ , we get $\cos^{2} ϕ = 1 - (1 - x^{2}) = x^{2}$ . Since $ϕ$ is in the first quadrant, $\cos ϕ = x$ .

\cot (\sin^{- 1} \sqrt{1 - x^{2}}) = \cot ϕ = \frac{\cos ϕ}{\sin ϕ} = \frac{x}{\sqrt{1 - x^{2}}}

Step 3: Equate both sides and solve for

x

○ Expand

Equating the simplified LHS and RHS:

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

Since $x \in (0, 1)$ , $x \neq 0$ , we can divide both sides by $x$ :

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

Square both sides:

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

Cross-multiply:

2 (1 - x^{2}) = 2 x^{2} + 1 ⟹ 2 - 2 x^{2} = 2 x^{2} + 1 ⟹ 1 = 4 x^{2} ⟹ x^{2} = \frac{1}{4}

. Since

x \in (0, 1)

, we take the positive root, so

x = \frac{1}{2}

💡 Teacher's Secret Hint

Always verify the solution against the given domain of $x$ to ensure validity.

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Maths Question 13 – JEE-MAIN 2026

Clue 1 of 3: Inverse Trigonometric Identities

🧠 Full Solution Path