JEE Main 2026 Apr 6 Shift 2, Mathematics Q13: Inverse Trigonometric Functions

JEE Main2026Apr 6, Shift 2Mathematics

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

A
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{2}{3}$
D
$\frac{5}{8}$

Solution

LHS: $sin (tan^{- 1} (x 2)) = \frac{x 2}{2 x ^{2} + 1}$ (right triangle with opposite $x 2$ , adjacent $1$ ).

RHS: $sin^{- 1} 1 - x^{2} = θ$ where $sin θ = 1 - x^{2}$ , so $cos θ = x$ and $cot θ = \frac{x}{1 - x ^{2}}$ .

Equating: $\frac{x 2}{2 x ^{2} + 1} = \frac{x}{1 - x ^{2}}$ . Cancel $x$ ( $x \neq = 0$ ) and square:

$\frac{2}{2 x ^{2} + 1} = \frac{1}{1 - x ^{2}}$ $\Rightarrow$ $2 (1 - x^{2}) = 2 x^{2} + 1$ $\Rightarrow$ $4 x^{2} = 1$ $\Rightarrow$ $x = \frac{1}{2}$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q13: Inverse Trigonometric Functions

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