Inverse Secant of Sum of Consecutive Shifted Secant Terms

To find the value of the given expression, let us define the angle: $\theta_k = \frac{7\pi}{12} + \frac{k\pi}{2}$

Notice that the difference between consecutive angles is: $\theta_{k+1} - \theta_k = \frac{\pi}{2}$

Since $\sin\left(\frac{\pi}{2}\right) = 1$, each term in the summation can be written as: $\sec\theta_k \sec\theta_{k+1} = \frac{1}{\cos\theta_k \cos\theta_{k+1}} = \frac{\sin(\theta_{k+1} - \theta_k)}{\cos\theta_k \cos\theta_{k+1}}$

Using the sine subtraction formula, $\sin(A - B) = \sin A \cos B - \cos A \sin B$: $\sec\theta_k \sec\theta_{k+1} = \frac{\sin\theta_{k+1}\cos\theta_k - \cos\theta_{k+1}\sin\theta_k}{\cos\theta_k \cos\theta_{k+1}} = \tan\theta_{k+1} - \tan\theta_k$

Now, substituting this back into the summation, we obtain a telescoping series: $S = \sum_{k=0}^{10} \sec\left(\frac{7\pi}{12}+\frac{k\pi}{2}\right)\sec\left(\frac{7\pi}{12}+\frac{(k+1)\pi}{2}\right) = \sum_{k=0}^{10} (\tan\theta_{k+1} - \tan\theta_k)$ $S = \tan\theta_{11} - \tan\theta_0$

Let us evaluate $\theta_0$ and $\theta_{11}$:

1. For $k=0$: $\theta_0 = \frac{7\pi}{12} = \frac{\pi}{2} + \frac{\pi}{12} \implies \tan\theta_0 = \tan\left(\frac{\pi}{2} + \frac{\pi}{12}\right) = -\cot\left(\frac{\pi}{12}\right)$

2. For $k=11$: $\theta_{11} = \frac{7\pi}{12} + \frac{11\pi}{2} = \frac{7\pi + 66\pi}{12} = \frac{73\pi}{12} = 6\pi + \frac{\pi}{12} \implies \tan\theta_{11} = \tan\left(6\pi + \frac{\pi}{12}\right) = \tan\left(\frac{\pi}{12}\right)$

Therefore: $S = \tan\left(\frac{\pi}{12}\right) - \left(-\cot\left(\frac{\pi}{12}\right)\right) = \tan\left(\frac{\pi}{12}\right) + \cot\left(\frac{\pi}{12}\right)$

Using the identity $\tan x + \cot x = \frac{2}{\sin 2x}$: $S = \frac{2}{\sin\left(2 \cdot \frac{\pi}{12}\right)} = \frac{2}{\sin\left(\frac{\pi}{6}\right)} = \frac{2}{\frac{1}{2}} = 4$

Now, substituting the value of $S$ into the given expression: $\sec^{-1}\left(\frac{1}{4} S\right) = \sec^{-1}\left(\frac{1}{4} \cdot 4\right) = \sec^{-1}(1)$

Since $\sec(0) = 1$ and $0 \in \left[-\frac{\pi}{4}, \frac{3\pi}{4}\right]$, the required value is: $\sec^{-1}(1) = 0$