JEE Main 2022 — Mathematics Trigonometry

Let $\theta +(m-1)\frac{\pi }{6}=x$θ+(m−1)6π​=x and $\theta +m\frac{\pi }{6}=y$θ+m6π​=y

So, $y-x=\frac{\pi }{6}$y−x=6π​

Now, $\displaystyle \sum _{m=1}^{9}\mathrm{sec}(\theta +(m-1)\frac{\pi }{6})\mathrm{sec}(\theta +\frac{m\pi }{6})$m=1∑9​sec(θ+(m−1)6π​)sec(θ+6mπ​)

$$=\sum _{m=1}^{9}\mathrm{sec}x\mathrm{sec}y=\sum _{m=1}^{9}\frac{1}{\mathrm{cos}x\mathrm{cos}y}$$=m=1∑9​secxsecy=m=1∑9​cosxcosy1​

$$=2\sum _{m=1}^{9}\frac{\mathrm{sin}(y-x)}{\mathrm{cos}x\mathrm{cos}y}=2\sum _{m=1}^{9}(\mathrm{tan}y-\mathrm{tan}x)$$=2m=1∑9​cosxcosysin(y−x)​=2m=1∑9​(tany−tanx)

$$=2\sum _{m=1}^{9}(\mathrm{tan}(\theta +m\frac{\pi }{6})-\mathrm{tan}(\theta +(m-1)\frac{\pi }{6}))$$=2m=1∑9​(tan(θ+m6π​)−tan(θ+(m−1)6π​))

$$=2(\mathrm{tan}(\theta +\frac{\pi }{6})-\mathrm{tan}(\theta +\frac{0\cdot \pi }{6}))+2(\mathrm{tan}(\theta +2\frac{\pi }{6})-\mathrm{tan}(\theta +\frac{\pi }{6}))$$=2(tan(θ+6π​)−tan(θ+60⋅π​))+2(tan(θ+26π​)−tan(θ+6π​))

$$+2(\mathrm{tan}(\theta +3\frac{\pi }{6})-\mathrm{tan}(\theta +2\frac{\pi }{6}))+...+2(\mathrm{tan}(\theta +9\frac{\pi }{6})-\mathrm{tan}(\theta +8\frac{\pi }{6}))$$+2(tan(θ+36π​)−tan(θ+26π​))+...+2(tan(θ+96π​)−tan(θ+86π​))

$$=2(\mathrm{tan}(\theta +\frac{9\pi }{6})-\mathrm{tan}\theta )=2(-\mathrm{cot}\theta -\mathrm{tan}\theta )$$=2(tan(θ+69π​)−tanθ)=2(−cotθ−tanθ)

i.e. $2(-\mathrm{cot}\theta -\mathrm{tan}\theta )=-\frac{8}{\sqrt{3}}$2(−cotθ−tanθ)=−3​8​ (Given)

$$\therefore \mathrm{tan}\theta +\mathrm{cot}\theta =\frac{4}{\sqrt{3}}$$∴tanθ+cotθ=3​4​

$\Rightarrow \mathrm{tan}\theta =\frac{1}{\sqrt{3}}$⇒tanθ=3​1​ or $\sqrt{3}$3​ (as $\theta \in (0,\frac{\pi }{2})$θ∈(0,2π​))

So, $S={\frac{\pi }{6},\frac{\pi }{3}}$S=6π​,3π​

Hence, $\displaystyle \underset{\theta \in S}{\sum }\theta =\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{2}$θ∈S∑​θ=6π​+3π​=2π​