CUET UG 2025 — Mathematics Algebra

The expression to evaluate is $\cos(2\cos^{-1}x + \sin^{-1}x)$cos(2cos−1x+sin−1x) at $x = \frac{1}{5}$x=51​.

Using the identity $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$cos−1x+sin−1x=2π​:

$$2\cos^{-1}x + \sin^{-1}x = \cos^{-1}x + \cos^{-1}x + \sin^{-1}x$$2cos−1x+sin−1x=cos−1x+cos−1x+sin−1x

$$= \cos^{-1}x + (\cos^{-1}x + \sin^{-1}x)$$=cos−1x+(cos−1x+sin−1x)

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

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Applying the cosine shift formula $\cos\left(A + \frac{\pi}{2}\right) = -\sin(A)$cos(A+2π​)=−sin(A):

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

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Let $\cos^{-1}x = \theta$cos−1x=θ, then $\cos\theta = x$cosθ=x.

Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1:

$$\sin\theta = \sqrt{1 - \cos^2\theta}$$sinθ=1−cos2θ​

$$= \sqrt{1 - x^2}$$=1−x2​

Therefore $\sin(\cos^{-1}x) = \sqrt{1-x^2}$sin(cos−1x)=1−x2​.

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The expression simplifies to:

$$\cos(2\cos^{-1}x + \sin^{-1}x) = -\sqrt{1-x^2}$$cos(2cos−1x+sin−1x)=−1−x2​

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Substituting $x = \frac{1}{5}$x=51​:

$$= -\sqrt{1 - \left(\frac{1}{5}\right)^2}$$=−1−(51​)2​

$$= -\sqrt{1 - \frac{1}{25}}$$=−1−251​​

$$= -\sqrt{\frac{24}{25}}$$=−2524​​

Therefore, the value is $-\sqrt{\frac{24}{25}}$−2524​​.