The value of sec - 1 ( 1 4 k = 0 10 sec ( 7 12 + k 2 ) sec ( 7 12 + ( k + 1 ) 2 ) ) in the interval [ - 4 , 3 4 ] equals _______ . [2019]

Question

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 _______ . [2019]

Smart Achievers · Accepted Answer

(0)

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

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

$= \sec^{- 1} [\frac{1}{2} \sum_{k = 0}^{10} \frac{- 1}{\sin (\frac{7 π}{6} + k π)}]$

$= \sec^{- 1} [\frac{1}{2} \sum_{k = 0}^{10} \frac{- 1}{\sin ((k + 1) π + \frac{π}{6})}]$

If $k$ is an even integer, then

$\sin ((k + 1) π + \frac{π}{6}) = - \sin \frac{π}{6} = \frac{- 1}{2}$

If $k$ is an odd integer, then $\sin ((k + 1) π + \frac{π}{6}) = \sin \frac{π}{6} = \frac{1}{2}$

$∴$ $\sum_{k = 0}^{9} \frac{1}{\sin ((k + 1) π + \frac{π}{6})} = 0$

Hence, $\sec^{- 1} [\frac{1}{2} \sum_{k = 0}^{10} \frac{- 1}{\sin ((k + 1) π + \frac{π}{6})}]$

$= \sec^{- 1} [\frac{1}{2} (\frac{- 1}{- \frac{1}{2}})] = \sec^{- 1} (1) = 0$

Solution

Q.
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 _______ . [2019]

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Solution

Q. The value of sec-1(14∑k=010sec(7π12+kπ2)sec(7π12+(k+1)π2)) in the interval [-π4,3π4] equals _______ . [2019]

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Q.
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 _______ . [2019]