JEE Main 2022 — Mathematics Trigonometry

Given, ${8}^{2{\mathrm{sin}}^{2}\theta }+{8}^{2{\mathrm{cos}}^{2}\theta }=16$82sin2θ+82cos2θ=16

$$\Rightarrow {8}^{2{\mathrm{sin}}^{2}\theta }+{8}^{2-2{\mathrm{sin}}^{2}\theta }=16$$⇒82sin2θ+82−2sin2θ=16

Now let ${8}^{2{\mathrm{sin}}^{2}\theta }=y$82sin2θ=y

$$\Rightarrow y+\frac{64}{y}=16$$⇒y+y64​=16

$$\Rightarrow y=8$$⇒y=8

$$\Rightarrow {8}^{2{\mathrm{sin}}^{2}\theta }=8$$⇒82sin2θ=8

$$\Rightarrow {\mathrm{sin}}^{2}\theta =\frac{1}{2}$$⇒sin2θ=21​

$$\Rightarrow \theta \in {\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}}$$⇒θ∈4π​,43π​,45π​,47π​

Now $\displaystyle n(S)+\underset{\theta \in S}{\sum }(\mathrm{sec}(\frac{\pi }{4}+2\theta )cosec(\frac{\pi }{4}+2\theta ))$n(S)+θ∈S∑​(sec(4π​+2θ)cosec(4π​+2θ))

$$=n(S)+\underset{\theta \in S}{\sum }\frac{1}{\mathrm{cos}(\frac{\pi }{4}+2\theta )\mathrm{sin}(\frac{\pi }{4}+2\theta )}$$=n(S)+θ∈S∑​cos(4π​+2θ)sin(4π​+2θ)1​

$$=4+\underset{\theta \in S}{\sum }\frac{2}{2\mathrm{cos}(\frac{\pi }{4}+2\theta )\mathrm{sin}(\frac{\pi }{4}+2\theta )}$$=4+θ∈S∑​2cos(4π​+2θ)sin(4π​+2θ)2​

$$=4+\underset{\theta \in S}{\sum }2\mathrm{cos}\text{ec}(\frac{\pi }{2}+4\theta )$$=4+θ∈S∑​2cosec(2π​+4θ)

$$=4+2\mathrm{cos}\text{ec}(\frac{\pi }{2}+\pi )\text{+}2\mathrm{cos}\text{ec}(\frac{\pi }{2}+3\pi )\text{+}2\mathrm{cos}\text{ec}(\frac{\pi }{2}+5\pi )\text{+}2\mathrm{cos}\text{ec}(\frac{\pi }{2}+7\pi )$$=4+2cosec(2π​+π)+2cosec(2π​+3π)+2cosec(2π​+5π)+2cosec(2π​+7π)

$$=4+2[-\mathrm{cos}\text{ec}\frac{\pi }{2}-\mathrm{cos}\text{ec}\frac{\pi }{2}-\mathrm{cos}\text{ec}\frac{\pi }{2}-\mathrm{cos}\text{ec}\frac{\pi }{2}]$$=4+2[−cosec2π​−cosec2π​−cosec2π​−cosec2π​]

$$=4+(-2)\times 4=-4$$=4+(−2)×4=−4