Question

What is the value of $\frac{(\cos {10}^o+\sin {20}^o)}{(\cos {20}^o-\sin {10}^o)}$?

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $-\sqrt{3}$

Answer 1

Answer: (C)

$\sqrt{3}$

$\cos {10}^{\circ }=\sin ({90}^{\circ }-{10}^{\circ })=\sin {80}^{\circ }$

$\cos {20}^{\circ }=\sin ({90}^{\circ }-{20}^{\circ })=\sin {70}^{\circ }$

$\frac{\cos {10}^{\circ }+\sin {20}^{\circ }}{\cos {20}^{\circ }-\sin {10}^{\circ }}=\frac{\sin {80}^{\circ }+\sin {20}^{\circ }}{\sin {70}^{\circ }-\sin {10}^{\circ }}=\frac{2\sin (\frac{{80}^{\circ }+{20}^{\circ }}{2})\cos (\frac{{80}^{\circ }-{20}^{\circ }}{2})}{2\cos (\frac{{70}^{\circ }+{10}^{\circ }}{2})\sin (\frac{{70}^{\circ }-{10}^{\circ }}{2})}=\frac{\sin {50}^{\circ }\cos {30}^{\circ }}{\cos {40}^{\circ }\sin {30}^{\circ }}=\cot {30}^{\circ }=\sqrt{3}$