CUET UG 2025 — Mathematics Geometry

Let $\theta = \sec^{-1} 2$θ=sec−12

This means $\sec \theta = 2$secθ=2

Since $\sec \theta = \frac{1}{\cos \theta}$secθ=cosθ1​:

$$\cos \theta = \frac{1}{2}$$cosθ=21​

Using the identity $\sec^2 \theta - \tan^2 \theta = 1$sec2θ−tan2θ=1:

$$\tan^2 \theta = \sec^2 \theta - 1$$tan2θ=sec2θ−1

$$\tan^2 \theta = (2)^2 - 1$$tan2θ=(2)2−1

$$\tan^2 \theta = 4 - 1$$tan2θ=4−1

$$\tan^2 \theta = 3$$tan2θ=3

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Let $\phi = \cosec^{-1} 3$ϕ=cosec−13

This means $\cosec \phi = 3$cosecϕ=3

Since $\cosec \phi = \frac{1}{\sin \phi}$cosecϕ=sinϕ1​:

$$\sin \phi = \frac{1}{3}$$sinϕ=31​

Using the identity $\cosec^2 \phi - \cot^2 \phi = 1$cosec2ϕ−cot2ϕ=1:

$$\cot^2 \phi = \cosec^2 \phi - 1$$cot2ϕ=cosec2ϕ−1

$$\cot^2 \phi = (3)^2 - 1$$cot2ϕ=(3)2−1

$$\cot^2 \phi = 9 - 1$$cot2ϕ=9−1

$$\cot^2 \phi = 8$$cot2ϕ=8

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$$\tan^2(\sec^{-1} 2) + \cot^2(\cosec^{-1} 3) = 3 + 8$$tan2(sec−12)+cot2(cosec−13)=3+8

$$= 11$$=11

Therefore, the value is $11$11.