CUET UG 2025 — Mathematics Geometry

Let $\theta = \cot^{-1}y$θ=cot−1y

This means $\cot\theta = y$cotθ=y

Using the identity $1 + \cot^2\theta = \csc^2\theta$1+cot2θ=csc2θ:

$$\csc^2\theta = 1 + \cot^2\theta$$csc2θ=1+cot2θ

$$\csc^2\theta = 1 + y^2$$csc2θ=1+y2

Therefore:

$$-\csc^2(\cot^{-1}y) = -(1 + y^2)$$−csc2(cot−1y)=−(1+y2)

$$-\csc^2(\cot^{-1}y) = -1 - y^2$$−csc2(cot−1y)=−1−y2

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Let $\alpha = \tan^{-1}x$α=tan−1x

This means $\tan\alpha = x$tanα=x

Using the identity $1 + \tan^2\alpha = \sec^2\alpha$1+tan2α=sec2α:

$$\sec^2\alpha = 1 + \tan^2\alpha$$sec2α=1+tan2α

$$\sec^2\alpha = 1 + x^2$$sec2α=1+x2

Therefore:

$$\sec^2(\tan^{-1}x) = 1 + x^2$$sec2(tan−1x)=1+x2

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Combining both terms:

$$-\csc^2(\cot^{-1}y) + \sec^2(\tan^{-1}x) = (-1 - y^2) + (1 + x^2)$$−csc2(cot−1y)+sec2(tan−1x)=(−1−y2)+(1+x2)

$$= -1 - y^2 + 1 + x^2$$=−1−y2+1+x2

$$= x^2 - y^2$$=x2−y2

Therefore, the value is $x^2 - y^2$x2−y2.