Question

The value of
$sec^{-1} \left(\frac{1}{4} \sum\limits^{10}_{k = 0} sec\left(\frac{7\pi}{12}+\frac{k\pi}{2}\right)sec\left(\frac{7\pi}{12}+\frac{\left(k+1\right)\pi}{2}\right)\right)$
in the interval $\left[-\frac{\pi}{4}, \frac{3\pi}{4}\right]$ equals

Answer 1

$sec^{-1}\left(\frac{1}{4}\displaystyle \sum_{k=0}^{10}\frac{1}{cos\left(\frac{7\pi}{12}+\frac{k\pi}{12}\right)cos\left(\frac{7\pi}{12}+\frac{\left(k+1\right)\pi}{2}\right)} \right)$
$=sec^{-1}\left(\frac{1}{4}\displaystyle \sum_{k=0}^{10} \frac{sin\left(\frac{7\pi}{12}+\frac{\left(k+1\right)\pi}{2}-\left(\frac{7\pi}{12}+\frac{k\pi}{2}\right)\right)}{cos\left(\frac{7\pi}{12}+\frac{k\pi}{2}\right).cos\left(\frac{7\pi}{12}+\frac{\left(k+1\right)\pi}{2}\right)}\right)$
$=sec^{-1}\left(\frac{1}{4}\left(\displaystyle \sum_{k=0}^{10}tan \left(\frac{7\pi}{12}+\left(k+1\right) \frac{\pi}{2}\right)-tan\left(\frac{7\pi}{12}+\frac{k\pi}{2}\right)\right)\right)$
$ =sec^{-1}\left(\frac{1}{4}\left(tan\left(\frac{11\pi}{2}+\frac{7\pi}{12}\right)-tan\left(\frac{7\pi}{12}\right)\right)\right)$
$=sec^{-1}\left(\frac{1}{4}\left(-cot \frac{7\pi}{12}-tan \frac{7\pi}{12}\right)\right)$
$=sec^{-1}\left(\frac{1}{4}\left(-\frac{1}{sin \frac{7\pi}{12}cos \frac{7\pi}{12}}\right)\right)$
$=sec^{-1}\left(-\frac{1}{2}\times\frac{1}{sin \frac{7\pi}{6}}\right)=sec^{-1}\left(1\right)=0.00$