Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ is equal to
MathematicsTrigonometric Ratios & IdentitiesTS EAMCETTS EAMCET 2021 (06 Aug Shift 1)
Options:
  • A 1
  • B 2
  • C 3
  • D 4
Solution:
1293 Upvotes Verified Answer
The correct answer is: 4
$\begin{aligned} & \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=4\left(\frac{\sqrt{3} \operatorname{cosec} 20^{\circ}}{4}-\frac{\sec 20^{\circ}}{4}\right) \\ & =4\left(\frac{\frac{\sqrt{3}}{2}}{2 \sin 20^{\circ}}-\frac{\frac{1}{2}}{2 \cos 20^{\circ}}\right)=4\left(\frac{\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}}{2 \sin 20^{\circ} \cdot \cos 20^{\circ}}\right) \\ & =4\left(\frac{\sin 60^{\circ} \cdot \cos 20^{\circ}-\cos 60^{\circ} \cdot \sin 20^{\circ}}{\sin 40^{\circ}}\right) \\ & =4 \frac{\sin \left(60^{\circ}-20^{\circ}\right)}{\sin 40^{\circ}}=4 \frac{\sin 40^{\circ}}{\sin 40^{\circ}}=4\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.