QJEEM22Ma16d6: JEE Main 2022 Mathematics - Class 11 JEEM22Ma16d6 Solution: Let Then is equal to:

$S = \left\{ {\theta \in [0,2\pi ]:{8^{2{{\sin }^2}\theta }} + {8^{2{{\cos }^2}\theta }} = 16} \right\}$

Now apply AM $\ge$ GM for ${8^{2{{\sin }^2}\theta }},\,{8^{2{{\cos }^2}\theta }}$

${{{8^{2{{\sin }^2}\theta }} + {8^{2{{\cos }^2}\theta }}} \over 2} \ge {\left( {{8^{2{{\sin }^2}\theta + 2{{\cos }^2}\theta }}} \right)^{{1 \over 2}}}$

$8 \ge 8$

$\Rightarrow {8^{2{{\sin }^2}\theta }} = {8^{2{{\cos }^2}\theta }}$

or ${\sin ^2}\theta = {\cos ^2}\theta$

$\therefore$ $\theta = {\pi \over 4},{{3\pi } \over 4},{{5\pi } \over 4},{{7\pi } \over 4}$

$n(S) + \sum\limits_{\theta \in S}^{} {\sec \left( {{\pi \over 4} + 2\theta } \right)\cos ec\left( {{\pi \over 4} + 2\theta } \right)}$

$4 + \sum\limits_{\theta \in S}^{} {{2 \over {2\sin \left( {{\pi \over 4} + 2\theta } \right)\cos \left( {{\pi \over 4} + 2\theta } \right)}}}$

$= 4 + \sum\limits_{\theta \in S}^{} {{2 \over {\sin \left( {{\pi \over 2} + 4\theta } \right)}} = 4 + 2\sum\limits_{\theta \in S}^{} {\cos ec\left( {{\pi \over 2} + 4\theta } \right)} }$

$= 4 + 2\left[ {\cos ec\left( {{\pi \over 2} + \pi } \right)\cos ec\left( {{\pi \over 2} + 3\pi } \right) + \cos ec\left( {{\pi \over 2} + 5\pi } \right) + \cos ec\left( {{\pi \over 2} + 7\pi } \right)} \right]$

$= 4 + 2\left[ { - \cos ec{\pi \over 2} - \cos ec{\pi \over 2} - \cos ec{\pi \over 2} - \cos ec{\pi \over 2}} \right]$

$= 4 - 2(4)$

$= 4 - 8$

$= - 4$