JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q13

MathematicsSingle correct+4 / -1Question 13 of 75

If $\sin\left(\tan^{-1}\left(x\sqrt2\right)\right)=\cot\left(\sin^{-1}\sqrt{1-x^2}\right)$ , $x\in(0,1)$ , then the value of $x$ is :

\dfrac12

\dfrac13

\dfrac23

\dfrac58

Answer and worked solution

Official answer: A

Left side. $\sin\left(\tan^{-1}x\sqrt2\right)=\dfrac{x\sqrt2}{\sqrt{1+2x^2}}$ .

Right side. With $\sin\theta=\sqrt{1-x^2}$ and $\cos\theta=x$ : $\cot\theta=\dfrac x{\sqrt{1-x^2}}$ .

Equate. $2(1-x^2)=1+2x^2$ , so $x^2=\tfrac14$ and $x=\tfrac12$ .

(A)

Topics:Inverse trigonometric functions

JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 13