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
$\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$ is equal to
MathematicsTrigonometric Ratios & IdentitiesKCETKCET 2012
Options:
  • A $\frac{1}{\sqrt{3}}$
  • B $\sqrt{3}$
  • C $\frac{1}{2}$
  • D 1
Solution:
1791 Upvotes Verified Answer
The correct answer is: $\sqrt{3}$
$\begin{aligned} \frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}} \\=& \frac{\sin \left(90^{\circ}-20^{\circ}\right)+\sin \left(90^{\circ}-40^{\circ}\right)}{\cos \left(90^{\circ}-20^{\circ}\right)+\sin \left(90^{\circ}-50^{\circ}\right)} \\=& \frac{\cos 20^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\cos 50^{\circ}} \\=& \frac{2 \cos 30^{\circ} \cos 10^{\circ}}{2 \cos 60^{\circ} \cos 10^{\circ}}=\frac{2 \cos 30^{\circ}}{2 \cos 60^{\circ}} \\=& \frac{2(\sqrt{3} / 2)}{2\left(\frac{1}{2}\right)}=\sqrt{3} \end{aligned}$

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.