Hard MCQ +4 / -1 PYQ · JEE Mains 2022

Let $$S=\left\{\theta \in[0,2 \pi]: 8^{2 \sin ^{2} \theta}+8^{2 \cos ^{2} \theta}=16\right\} .$$ Then $$n(s) + \sum\limits_{\theta \in S}^{} {\left( {\sec \left( {{\pi \over 4} + 2\theta } \right)\cos ec\left( {{\pi \over 4} + 2\theta } \right)} \right)} $$ is equal to:

  1. A 0
  2. B $-$2
  3. C $-$4 Correct answer
  4. D 12

Solution

<p>$$S = \left\{ {\theta \in [0,2\pi ]:{8^{2{{\sin }^2}\theta }} + {8^{2{{\cos }^2}\theta }} = 16} \right\}$$</p> <p>Now apply AM $\ge$ GM for ${8^{2{{\sin }^2}\theta }},\,{8^{2{{\cos }^2}\theta }}$</p> <p>$${{{8^{2{{\sin }^2}\theta }} + {8^{2{{\cos }^2}\theta }}} \over 2} \ge {\left( {{8^{2{{\sin }^2}\theta + 2{{\cos }^2}\theta }}} \right)^{{1 \over 2}}}$$</p> <p>$8 \ge 8$</p> <p>$\Rightarrow {8^{2{{\sin }^2}\theta }} = {8^{2{{\cos }^2}\theta }}$</p> <p>or ${\sin ^2}\theta = {\cos ^2}\theta$</p> <p>$\therefore$ $\theta = {\pi \over 4},{{3\pi } \over 4},{{5\pi } \over 4},{{7\pi } \over 4}$</p> <p>$$n(S) + \sum\limits_{\theta \in S}^{} {\sec \left( {{\pi \over 4} + 2\theta } \right)\cos ec\left( {{\pi \over 4} + 2\theta } \right)} $$</p> <p>$$4 + \sum\limits_{\theta \in S}^{} {{2 \over {2\sin \left( {{\pi \over 4} + 2\theta } \right)\cos \left( {{\pi \over 4} + 2\theta } \right)}}} $$</p> <p>$$ = 4 + \sum\limits_{\theta \in S}^{} {{2 \over {\sin \left( {{\pi \over 2} + 4\theta } \right)}} = 4 + 2\sum\limits_{\theta \in S}^{} {\cos ec\left( {{\pi \over 2} + 4\theta } \right)} } $$</p> <p>$$ = 4 + 2\left[ {\cos ec\left( {{\pi \over 2} + \pi } \right)\cos ec\left( {{\pi \over 2} + 3\pi } \right) + \cos ec\left( {{\pi \over 2} + 5\pi } \right) + \cos ec\left( {{\pi \over 2} + 7\pi } \right)} \right]$$</p> <p>$$ = 4 + 2\left[ { - \cos ec{\pi \over 2} - \cos ec{\pi \over 2} - \cos ec{\pi \over 2} - \cos ec{\pi \over 2}} \right]$$</p> <p>$= 4 - 2(4)$</p> <p>$= 4 - 8$</p> <p>$= - 4$</p>

About this question

Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Ratios and Identities

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