Search any question & find its solution
Question:
Answered & Verified by Expert
$\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}}=$
Options:
Solution:
2202 Upvotes
Verified Answer
The correct answer is:
$\frac{4}{\sqrt{3}}$
$\begin{aligned} & \text {Since } \frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}} \\ & =\frac{1}{\cos \left(270+20^{\circ}\right)}+\frac{1}{\sqrt{3} \sin \left(270-20^{\circ}\right)}\end{aligned}$
$\begin{aligned} & =\frac{1}{\sin 20^{\circ}}-\frac{1}{\sqrt{3} \sin 20^{\circ}}=\frac{2\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} \\ & =\frac{2 \sin \left(60^{\circ}-20^{\circ}\right)}{\frac{\sqrt{3}}{2} \sin 40^{\circ}}=\frac{4}{\sqrt{3}}\end{aligned}$
$\begin{aligned} & =\frac{1}{\sin 20^{\circ}}-\frac{1}{\sqrt{3} \sin 20^{\circ}}=\frac{2\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} \\ & =\frac{2 \sin \left(60^{\circ}-20^{\circ}\right)}{\frac{\sqrt{3}}{2} \sin 40^{\circ}}=\frac{4}{\sqrt{3}}\end{aligned}$
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.