The value of sec - 1 ⁡ 1 4   ∑ k = 0 10 sec ⁡ 7 π 12 + k π 2 sec ⁡ 7…

The value of sec-1⁡14 ∑k=010sec⁡7π12+kπ2sec⁡7π12+k+1π2 in the interval -π4,3π4 equals

Solution

sec −1 1 4 ∑ k=0 10 sec 7π 12 + kπ 2 sec 7π 12 + kπ 2 + π 2
= sec −1 −1 4 ∑ k=0 10 sec 7π 12 + kπ 2 cosec 7π 12 + kπ 2 ∴sec π 2 +θ =−cosecθ
=sec-1-14∑k=0101cos⁡7π12+kπ2sin⁡7π12+kπ2
=sec-1⁡-14∑k=0102sin⁡7π6+kπ∴2sin⁡θcos⁡θ=sin⁡2θ
now if k=evensin⁡2nπ+7π6=sin⁡7π6=-12
and if k=oddsin⁡2n+1π+7π6=-sin⁡7π6=12
hence
=sec-1⁡-121-12+112+1-12+…..
=sec-1⁡1
=0

Asked in: JEE Advanced 2019 (Paper 2)

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