JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 14)
The value of
$${\sec ^{ - 1}}\left( \matrix{ {1 \over 4}\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) \hfill \cr} \right)$$
in the interval $$\left[ { - {\pi \over 4},\,{{3\pi } \over 4}} \right]$$ equals ..........
$${\sec ^{ - 1}}\left( \matrix{ {1 \over 4}\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) \hfill \cr} \right)$$
in the interval $$\left[ { - {\pi \over 4},\,{{3\pi } \over 4}} \right]$$ equals ..........
Answer
0
Explanation
$$\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)$$
= $$\sum\limits_{k = 0}^{10} {{1 \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)}}} $$
= $$\sum\limits_{k = 0}^{10} {{{\sin \left[ {\left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) - \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \right]} \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)}}} $$
$$ \because $$ $$\left[ {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2} - \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right) = {\pi \over 2}\,and\,sin{\pi \over 2} = 1} \right]$$
$$\sum\limits_{k = 0}^{10} {{\matrix{ \sin \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right) \hfill \cr - \sin \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right) \hfill \cr} \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right)}}} $$
= $$\sum\limits_{k = 0}^{10} {\left[ {\tan \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \right]} $$
= $$\tan \left( {{{7\pi } \over {12}} + {\pi \over 2}} \right) - \tan \left( {{{7\pi } \over {12}}} \right) + \tan \left( {{{7\pi } \over {12}} + {{2\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {\pi \over 2}} \right)$$ ....$$ + \tan \left( {{{7\pi } \over {12}} + {{11\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {{10\pi } \over 2}} \right)$$
= $$\tan \left( {{{7\pi } \over {12}} + {{11\pi } \over 2}} \right) - \tan {{7\pi } \over {12}} = \tan {{7\pi } \over {12}} + \cot {\pi \over {12}}$$
= $${1 \over {\sin {\pi \over {12}}\cos {\pi \over {12}}}} = {2 \over {\sin {\pi \over 6}}} = 4$$
So, $${\sec ^{ - 1}}\left( {{1 \over 4}\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)} } \right)$$
= $${\sec ^{ - 1}}$$ (1) = 0
= $$\sum\limits_{k = 0}^{10} {{1 \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)}}} $$
= $$\sum\limits_{k = 0}^{10} {{{\sin \left[ {\left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) - \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \right]} \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)}}} $$
$$ \because $$ $$\left[ {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2} - \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right) = {\pi \over 2}\,and\,sin{\pi \over 2} = 1} \right]$$
$$\sum\limits_{k = 0}^{10} {{\matrix{ \sin \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right) \hfill \cr - \sin \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right) \hfill \cr} \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right)}}} $$
= $$\sum\limits_{k = 0}^{10} {\left[ {\tan \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \right]} $$
= $$\tan \left( {{{7\pi } \over {12}} + {\pi \over 2}} \right) - \tan \left( {{{7\pi } \over {12}}} \right) + \tan \left( {{{7\pi } \over {12}} + {{2\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {\pi \over 2}} \right)$$ ....$$ + \tan \left( {{{7\pi } \over {12}} + {{11\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {{10\pi } \over 2}} \right)$$
= $$\tan \left( {{{7\pi } \over {12}} + {{11\pi } \over 2}} \right) - \tan {{7\pi } \over {12}} = \tan {{7\pi } \over {12}} + \cot {\pi \over {12}}$$
= $${1 \over {\sin {\pi \over {12}}\cos {\pi \over {12}}}} = {2 \over {\sin {\pi \over 6}}} = 4$$
So, $${\sec ^{ - 1}}\left( {{1 \over 4}\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)} } \right)$$
= $${\sec ^{ - 1}}$$ (1) = 0
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