JEE Main 2022 · 27 Jul · Shift 2 · Mathematics Q19

← JEE Main 2022 · 27 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 19 of 90

Let $S=\left\{\theta\in\left(0,\dfrac\pi2\right):\displaystyle\sum_{m=1}^9\sec\left(\theta+(m-1)\frac\pi6\right)\sec\left(\theta+\frac{m\pi}6\right)=-\frac8{\sqrt3}\right\}$ . Then

S=\left\{\dfrac\pi{12}\right\}

S=\left\{\dfrac{2\pi}3\right\}

\displaystyle\sum_{\theta\in S}\theta=\frac\pi2

\displaystyle\sum_{\theta\in S}\theta=\frac{3\pi}4

Answer and worked solution

Official answer: C

Telescope. $\sec a\sec b=\frac{\tan b-\tan a}{\sin(b-a)}$ with $b-a=\frac\pi6$ . The sum is $2\left(\tan\left(\theta+\frac{3\pi}2\right)-\tan\theta\right)=-2(\cot\theta+\tan\theta)$ .

Solve. $\frac2{\sin2\theta}=\frac4{\sqrt3}$ , so $\theta=\frac\pi6,\frac\pi3$ , with sum $\frac\pi2$ .

(C)

Topics:Trigonometric equations Trigonometric ratios and identities

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JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 19