Question

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

Answer 1

Answer: 0

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