Q61 / 160 · AP EAMCET 2023 (15 May Shift 2) · Mathematics

$\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}}=$

  • A
    $\frac{\sqrt{3}}{4}$
  • Correct answer
    B
    $\frac{4}{\sqrt{3}}$
  • C
    $\frac{2}{\sqrt{3}}$
  • D
    $\frac{\sqrt{3}}{2}$

Solution

$\begin{aligned} & \text {Since } \frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}} \\ & =\frac{1}{\cos \left(270+20^{\circ}\right)}+\frac{1}{\sqrt{3} \sin \left(270-20^{\circ}\right)}\end{aligned}$ $\begin{aligned} & =\frac{1}{\sin 20^{\circ}}-\frac{1}{\sqrt{3} \sin 20^{\circ}}=\frac{2\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} \\ & =\frac{2 \sin \left(60^{\circ}-20^{\circ}\right)}{\frac{\sqrt{3}}{2} \sin 40^{\circ}}=\frac{4}{\sqrt{3}}\end{aligned}$

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

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