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JEE MainMathematicsToughInverse Trigonometric functions

The value of
sec⁡−1(14∑k=010\mspace2musec(7π12+kπ2)sec(7π12+(k+1)π2))\sec^{- 1}\left( \frac{1}{4}\sum_{k = 0}^{10}\mspace{2mu} 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
theinterval [−π4,3π4]\left\lbrack - \frac{\pi}{4},\frac{3\pi}{4} \right\rbrack equals

Show answer and solution
Evaluate
 sec⁡−1[14∑k=010\mspace2musec(7π12+kπ12)sec(7π12+(k+1)π2)]\sec^{- 1}\left\lbrack \frac{1}{4}\sum_{k = 0}^{10}\mspace{2mu} sec\left( \frac{7\pi}{12} + \frac{k\pi}{12} \right)sec\left( \frac{7\pi}{12} + (k + 1)\frac{\pi}{2} \right) \right\rbrack
Given exp
=sec⁡−1[−14∑k=010\mspace2musec(7π12+kπ12)cosec(7π12+kπ2)]= \sec^{- 1}\left\lbrack - \frac{1}{4}\sum_{k = 0}^{10}\mspace{2mu} sec\left( \frac{7\pi}{12} + \frac{k\pi}{12} \right)cosec\left( \frac{7\pi}{12} + \frac{k\pi}{2} \right) \right\rbrack
=sec⁡−1[−14∑k=010\mspace2mu2sin(7π6+kπ)]=sec⁡−1[−12∑k=010\mspace2mu1(−1)k+1sinπ6]= \sec^{- 1}\left\lbrack - \frac{1}{4}\sum_{k = 0}^{10}\mspace{2mu}\frac{2}{sin\left( \frac{7\pi}{6} + k\pi \right)} \right\rbrack = \sec^{- 1}\left\lbrack - \frac{1}{2}\sum_{k = 0}^{10}\mspace{2mu}\frac{1}{( - 1)^{k + 1}sin\frac{\pi}{6}} \right\rbrack
=sec⁡−1(−∑k=010\mspace2mu1(−1)k+1)=sec⁡−1(1)=0= \sec^{- 1}\left( - \sum_{k = 0}^{10}\mspace{2mu}\frac{1}{( - 1)^{k + 1}} \right) = \sec^{- 1}(1) = 0

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx

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