The value of sec ^{-1}left(frac{1}{4} sum_{k= | vedclass

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(A) Let $S = \sum_{k=0}^{10} \sec \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right) \sec \left(\frac{7 \pi}{12}+\frac{(k+1) \pi}{2}\right)$.Using the iden

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(A) Let $S = \sum_{k=0}^{10} \sec \left(\frac{7 \pi}{12}+\frac{k \pi}{2}ight) \sec \left(\frac{7 \pi}{12}+\frac{(k+1) \pi}{2}ight)$.Using the identity $\sec A \sec B = \frac{\sin(B-A)}{\cos A \cos B \sin(B-A)}$,we note that $B-A = \frac{\pi}{2}$.Thus,$\sec A \sec B = \frac{\sin(\pi/2)}{\cos A \cos B \sin(\pi/2)} = \frac{	an B - 	an A}{\sin(\pi/2)} = 	an B - 	an A$.Here,$A = \frac{7 \pi}{12} + \frac{k \pi}{2}$ and $B = \frac{7 \pi}{12} + \frac{(k+1) \pi}{2}$.So,the sum becomes $\sum_{k=0}^{10} (	an(\frac{7 \pi}{12} + \frac{(k+1) \pi}{2}) - 	an(\frac{7 \pi}{12} + \frac{k \pi}{2}))$.This is a telescoping sum: $	an(\frac{7 \pi}{12} + \frac{11 \pi}{2}) - 	an(\frac{7 \pi}{12})$.Since $	an(	heta + \frac{11 \pi}{2}) = 	an(	heta - \frac{\pi}{2}) = -\cot 	heta$,the sum is $-\cot(\frac{7 \pi}{12}) - 	an(\frac{7 \pi}{12}) = -(\frac{\cos(7 \pi / 12)}{\sin(7 \pi / 12)} + \frac{\sin(7 \pi / 12)}{\cos(7 \pi / 12)}) = -\frac{1}{\sin(7 \pi / 12) \cos(7 \pi / 12)} = -\frac{2}{\sin(7 \pi / 6)} = -\frac{2}{-1/2} = 4$.The expression is $\sec^{-1}(\frac{1}{4} 	imes 4) = \sec^{-1}(1) = 0$.

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

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