The value of sec−1⁡14∑k=010sec⁡7π12+kπ2sec⁡7π12+(k+1)π2 in the interval −π4,3π4 equals ____

sec−1⁡(14∑k=010sec⁡(7π12+kπ2)sec⁡(7π12+(k+1)π2))Sec−1⁡14∑k=010 1cos⁡7π12+kπ2cos⁡7π12+(k+1)π2⇒Sec−1⁡14∑k=010 sin⁡7π12+(k+1)π2−7π12+kπ2cos⁡7π12+kπ2cos⁡7π12+(k+1)π2Sec−1⁡14∑k=010 tan⁡7π12+(k+1)π2−tan⁡7π2+kπ2Sec−1⁡14 tan⁡7π12+π2−tan⁡7π12+tan⁡7π12+2π2−tan⁡7π12+π2…tan⁡7π12+11π2−tan⁡7π2+10π2Sec−1⁡14tan⁡7π12+11π2−tan⁡7π12⇒ Sec-114tanπ12-tan7π12⇒Sec-114((2-3)+2+3)Sec−1⁡(1)=0.00

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

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Detailed Solution

$\sec^{- 1} (\frac{1}{4} \sum_{k = 0}^{10} \sec (\frac{7 π}{12} + \frac{k π}{2}) \sec (\frac{7 π}{12} + \frac{(k + 1) π}{2}))$

${Sec}^{- 1} (\frac{1}{4} \sum_{k = 0}^{10} \frac{1}{\cos (\frac{7 π}{12} + \frac{k π}{2}) \cos (\frac{7 π}{12} + \frac{(k + 1) π}{2})})$

$\Rightarrow {Sec}^{- 1} (\frac{1}{4} \sum_{k = 0}^{10} \frac{\sin (\frac{7 π}{12} + \frac{(k + 1) π}{2}) - (\frac{7 π}{12} + \frac{k π}{2})}{\cos (\frac{7 π}{12} + \frac{k π}{2}) \cos (\frac{7 π}{12} + (k + 1) \frac{π}{2})})$

${Sec}^{- 1} (\frac{1}{4} \sum_{k = 0}^{10} (\tan (\frac{7 π}{12} + (k + 1) \frac{π}{2}) - \tan (\frac{7 π}{2} + k \frac{π}{2})))$

${Sec}^{- 1} (\frac{1}{4} (\tan (\frac{7 π}{12} + \frac{π}{2}) - \tan \frac{7 π}{12}) + (\tan (\frac{7 π}{12} + \frac{2 π}{2}) - \tan (\frac{7 π}{12} + \frac{π}{2})) \dots (\tan (\frac{7 π}{12} + \frac{11 π}{2})) - \tan (\frac{7 π}{2} + \frac{10 π}{2}))$