CUET UG 2025 — Mathematics Geometry

Given: $4\sin^{-1}x + \cos^{-1}x = \pi$4sin−1x+cos−1x=π where $x \in [-1,1]$x∈[−1,1]

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The fundamental identity states:

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$sin−1x+cos−1x=2π​ for all $x \in [-1,1]$x∈[−1,1]

This works because if an angle has sine = $x$x, then its complementary angle has cosine = $x$x, and they add up to $\frac{\pi}{2}$2π​ radians.

From this identity:

$$\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$$cos−1x=2π​−sin−1x

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Substituting into the given equation:

$$4\sin^{-1}x + \left(\frac{\pi}{2} - \sin^{-1}x\right) = \pi$$4sin−1x+(2π​−sin−1x)=π

$$4\sin^{-1}x - \sin^{-1}x + \frac{\pi}{2} = \pi$$4sin−1x−sin−1x+2π​=π

$$3\sin^{-1}x + \frac{\pi}{2} = \pi$$3sin−1x+2π​=π

$$3\sin^{-1}x = \pi - \frac{\pi}{2}$$3sin−1x=π−2π​

$$3\sin^{-1}x = \frac{\pi}{2}$$3sin−1x=2π​

$$\sin^{-1}x = \frac{\pi}{6}$$sin−1x=6π​

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Taking sine on both sides:

$$x = \sin\left(\frac{\pi}{6}\right)$$x=sin(6π​)

$$x = \frac{1}{2}$$x=21​

Therefore, $x = \dfrac{1}{2}$x=21​