Q88 / 150 · MHT CET 2024 (04 May Shift 1) · Mathematics

The value of the expression $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ is equal to

  • A
    2
  • B
    $\frac{2 \sin 20^{\circ}}{\sin 40^{\circ}}$
  • Correct answer
    C
    4
  • D
    $4 \frac{\sin 20^{\circ}}{\sin 40^{\circ}}$

Solution

$\begin{aligned} & \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ} \\ & =\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}} \\ & =\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \\ & =\frac{4\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)}{2 \sin 20^{\circ} \cos 20^{\circ}} \\ & =\frac{4\left(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}\right)}{\sin 40^{\circ}} \\ & =4 \frac{\sin 40^{\circ}}{\sin 40^{\circ}}=4\end{aligned}$

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

Relevant Questions