Q64 / 160 · AP EAMCET 2008 · Mathematics

$\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ is equal to

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

Solution

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

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

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