JEE Main 2022 — Trigonometric Ratios & Identities Question with Solution

Let $\theta +\left(m-1\right)\frac{\pi }{6}=x$ and $\theta +m\frac{\pi }{6}=y$

So, $y-x=\frac{\pi }{6}$

Now, $\sum _{m=1}^9\sec \left(\theta +\left(m-1\right)\frac{\pi }{6}\right)\sec \left(\theta +\frac{m\pi }{6}\right)$

$=\sum _{m=1}^9\sec x\sec y=\sum _{m=1}^9\frac{1}{\cos x\cos y}$

$=2\sum _{m=1}^9\frac{\sin \left(y-x\right)}{\cos x\cos y}=2\sum _{m=1}^9\left(\tan y-\tan x\right)$

$=2\sum _{m=1}^9\left(\tan \left(\theta +m\frac{\pi }{6}\right)-\tan \left(\theta +(m-1)\frac{\pi }{6}\right)\right)$

$=2\left(\tan \left(\theta +\frac{\pi }{6}\right)-\tan \left(\theta +\frac{0·\pi }{6}\right)\right)+2\left(\tan \left(\theta +2\frac{\pi }{6}\right)-\tan \left(\theta +\frac{\pi }{6}\right)\right)$

$+2\left(\tan \left(\theta +3\frac{\pi }{6}\right)-\tan \left(\theta +2\frac{\pi }{6}\right)\right)+ ... +2\left(\tan \left(\theta +9\frac{\pi }{6}\right)-\tan \left(\theta +8\frac{\pi }{6}\right)\right)$

$=2\left(\tan \left(\theta +\frac{9\pi }{6}\right)-\tan \theta \right)=2\left(-\cot \theta -\tan \theta \right)$

i.e. $2\left(-\cot \theta -\tan \theta \right)=-\frac{8}{\sqrt{3}}$ (Given)

$\therefore \tan \theta +\cot \theta =\frac{4}{\sqrt{3}}$

$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$ or $\sqrt{3}$ (as $\theta \in \left(0,\frac{\pi }{2}\right)$)

So, $S=\left\{\frac{\pi }{6},\frac{\pi }{3}\right\}$

Hence, $\sum _{\theta \in S}\theta =\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{2}$