Question bank/JEE Advanced/Mathematics

JEE AdvancedMathematicsChallengingJEE Adv 2019 Paper 2

What is the value of

sec⁡−1(14∑k=010sec⁡(7π12+kπ2)sec⁡(7π12+(k+1)π2))\sec ^{-1}\left(\frac{1}{4} \sum_{k=0}^{10} \sec \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right) \sec \left(\frac{7 \pi}{12}+\frac{(k+1) \pi}{2}\right)\right)

in the interval
[−π4,3π4]\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right]?

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Source: JEEBench (IIT Delhi) / JEE Adv 2019 Paper 2

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