The value of √3 20^ - 20^ 20^ 40^ 60^ 80^ is equal to

Question

The value of $\frac{\sqrt{3}\csc {20}^{\circ }-\sec {20}^{\circ }}{\cos {20}^{\circ }\cos {40}^{\circ }\cos {60}^{\circ }\cos {80}^{\circ }}$ is equal to

Accepted Answer

$\begin{gathered} \frac{\sqrt{3}\csc {20}^{\circ }-\sec {20}^{\circ }}{\cos {20}^{\circ }\cos {40}^{\circ }\cos {60}^{\circ }\cos {80}^{\circ }} \ =\frac{\frac{\sqrt{3}}{\sin {20}^{\circ }}-\frac{1}{\cos {20}^{\circ }}}{\cos {20}^{\circ }\cos {40}^{\circ }\cos {80}^{\circ }\cdot \frac{1}{2}},\, {\cos {60}^{\circ }=\frac{1}{2}} \ =\frac{2(\frac{\sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ }}{\sin {20}^{\circ }\cos {20}^{\circ }})}{\cos {20}^{\circ }\cos {40}^{\circ }\cos {80}^{\circ }},\, \ =\frac{2(\sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ })}{\frac{\sin {40}^{\circ }}{2}\cdot \cos {20}^{\circ }\cos {40}^{\circ }\cos {80}^{\circ }},{\sin {20}^{\circ }\cos {20}^{\circ }=\frac{\sin {40}^{\circ }}{2}} \ \sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ }=2\cdot (\frac{\sqrt{3}}{2}\cos {20}^{\circ }-\frac{1}{2}\sin {20}^{\circ }) \ =2(\sin {60}^{\circ }\cos {20}^{\circ }-\cos {60}^{\circ }\sin {20}^{\circ })=2\sin (60-20)=2\sin {40}^{\circ } \ \ =\frac{2\cdot 2\sin {40}^{\circ }}{\frac{\sin {80}^{\circ }}{4}\cos {20}^{\circ }\cos {80}^{\circ }}=\frac{4\sin {40}^{\circ }}{\frac{\sin {160}^{\circ }}{8}\cos {20}^{\circ }},{\sin {80}^{\circ }\cos {80}^{\circ }=\frac{\sin {160}^{\circ }}{2}} \ =\frac{4	imes 2\sin {20}^{\circ }\cdot \cos {20}^{\circ }}{\sin (180-160)\cos {20}^{\circ }}	imes 8=\frac{64\sin {20}^{\circ }}{\sin {20}^{\circ }}=64 \end{gathered}$

Solutions · 1