JEE Main 2026 — Mathematics Trigonometry

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

From the right-angled triangle, $\sin\alpha = \dfrac{x\sqrt{2}}{\sqrt{1+2x^2}}$sinα=1+2x2​x2​​

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

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

$$\cot\beta = \dfrac{\cos\beta}{\sin\beta} = \dfrac{x}{\sqrt{1-x^2}}$$cotβ=sinβcosβ​=1−x2​x​

Given $\sin\alpha = \cot\beta$sinα=cotβ, substituting the values:

$$\dfrac{x\sqrt{2}}{\sqrt{1+2x^2}} = \dfrac{x}{\sqrt{1-x^2}}$$1+2x2​x2​​=1−x2​x​

Since $x \in (0,1)$x∈(0,1), $x \neq 0$x=0. Dividing by $x$x and squaring both sides:

$$\dfrac{2}{1+2x^2} = \dfrac{1}{1-x^2}$$1+2x22​=1−x21​

$$2(1-x^2) = 1+2x^2$$2(1−x2)=1+2x2

$$2 - 2x^2 = 1 + 2x^2$$2−2x2=1+2x2

$$4x^2 = 1$$4x2=1

$$x^2 = \dfrac{1}{4}$$x2=41​

Since $x \in (0,1)$x∈(0,1), $x = \dfrac{1}{2}$x=21​

Answer: $\dfrac{1}{2}$21​