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

$\begin{array}{l}{\mathrm{Sec}}^{-1}\left[\frac{1}{4}\sum _{\mathrm{k}=0}^{10} \mathrm{Sec}\left(\frac{7\mathrm{\pi }}{12}+\frac{\mathrm{k\pi }}{2}\right)\mathrm{Sec}\left(\frac{7\mathrm{\pi }}{12}+\frac{(\mathrm{k}+1)\mathrm{\pi }}{2}\right)\right]\\ ={\mathrm{Sec}}^{-1}\left[\frac{1}{\frac{1}{2}}\sum _{\mathrm{k}=0}^{10} \frac{1}{2\cos \left(\frac{7\mathrm{\pi }}{12}+\frac{\mathrm{k\pi }}{2}\right)\mathrm{Cos}\left(\frac{7\mathrm{\pi }}{12}+\frac{\mathrm{k\pi }}{2}+\frac{\mathrm{\pi }}{2}\right)}\right]\\ ={\mathrm{Sec}}^{-1}\left[\frac{1}{2}\sum _{\mathrm{k}=0}^{10} \frac{-1}{\sin \left(\frac{7\mathrm{\pi }}{6}+\mathrm{k\pi }\right)}\right]\\ ={\mathrm{Sec}}^{-1}\left[\frac{1}{2}\sum _{\mathrm{k}=0}^{10} \frac{-1}{\sin \left(\right(\mathrm{K}+1)\mathrm{\pi }+\mathrm{\pi }/6)}\right]\end{array}$
If K is even then $\mathrm{Sin}\left((\mathrm{k}+1)\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)=-\sin \frac{\mathrm{\pi }}{6}=-\frac{1}{2}$
If K is odd then Sin $\mathrm{Sin}\left(\right(\mathrm{k}+1)\mathrm{\pi }+\mathrm{\pi }/6)=\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}$
$$\begin{gathered} \begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sum _{\mathrm{k}=0}^9 \frac{1}{\mathrm{Sin}\left((\mathrm{k}+1)\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)}=0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{Sec}}^{-1}\left[\frac{1}{2}\sum _{\mathrm{k}=0}^{10} \frac{-1}{\sin \left((\mathrm{k}+1)\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)}\right]\end{array} \\ {\sec }^{-1}\left(\frac{1}{2}\left(\frac{-1}{\frac{-1}{2}}\right)\right)={\mathrm{Sec}}^{-1}\left(1\right)=0 \end{gathered}$$