Q85 / 150 · MHT CET 2024 (11 May Shift 1) · Mathematics

If angle $\theta$ in $[0,2 \pi]$ satisfies both the equations $\cot \theta=\sqrt{3}$ and $\sqrt{3} \sec \theta+2=0$, then $\theta$ is equal to

  • A
    $\frac{\pi}{6}$
  • Correct answer
    B
    $\frac{7 \pi}{6}$
  • C
    $\frac{5 \pi}{6}$
  • D
    $\frac{11 \pi}{6}$

Solution

$\cot \theta=\sqrt{3}$ and $\sec \theta=\frac{-2}{\sqrt{3}}$ i.e., $\tan \theta=\frac{1}{\sqrt{3}}$ and $\cos \theta=-\frac{\sqrt{3}}{2}$ $\therefore \quad \theta$ lies in $3^{\text {rd }}$ quadrant $\therefore \quad \theta=\frac{7 \pi}{6}$

Hence, the correct answer is option: 'B'.

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