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JEE Main 20266 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2026 · 6 April · Evening Shift · Mathematics · Inverse Trigonometric Functions

If ( ^ -1 (x 2 )) = ( ^ -1 1-x^2 ) , x ∈ (0,1) , then the value of...

If sin⁡(tan⁡−1(x2))=cot⁡(sin⁡−11−x2)\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1-x^2}), x∈(0,1)x \in (0,1), then the value of xx is :

Options

  1. A12\dfrac{1}{2}
  2. B13\dfrac{1}{3}
  3. C23\dfrac{2}{3}
  4. D58\dfrac{5}{8}

Correct answer

A. 12\dfrac{1}{2}

Step-by-step solution

Let α=tan⁡−1(x2)⇒tan⁡α=x2\alpha = \tan^{-1}(x\sqrt{2}) \Rightarrow \tan\alpha = x\sqrt{2}

From the right-angled triangle, sin⁡α=x21+2x2\sin\alpha = \dfrac{x\sqrt{2}}{\sqrt{1+2x^2}}

Let β=sin⁡−11−x2⇒sin⁡β=1−x2\beta = \sin^{-1}\sqrt{1-x^2} \Rightarrow \sin\beta = \sqrt{1-x^2}

From the right-angled triangle, cos⁡β=1−(1−x2)=x\cos\beta = \sqrt{1 - (1-x^2)} = x

cot⁡β=cos⁡βsin⁡β=x1−x2\cot\beta = \dfrac{\cos\beta}{\sin\beta} = \dfrac{x}{\sqrt{1-x^2}}

Given sin⁡α=cot⁡β\sin\alpha = \cot\beta, substituting the values:

x21+2x2=x1−x2\dfrac{x\sqrt{2}}{\sqrt{1+2x^2}} = \dfrac{x}{\sqrt{1-x^2}}

Since x∈(0,1)x \in (0,1), x≠0x \neq 0. Dividing by xx and squaring both sides:

21+2x2=11−x2\dfrac{2}{1+2x^2} = \dfrac{1}{1-x^2}

2(1−x2)=1+2x22(1-x^2) = 1+2x^2

2−2x2=1+2x22 - 2x^2 = 1 + 2x^2

4x2=14x^2 = 1

x2=14x^2 = \dfrac{1}{4}

Since x∈(0,1)x \in (0,1), x=12x = \dfrac{1}{2}

Answer: 12\dfrac{1}{2}
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