Question

If $\frac{1}{\sin {20}^o}+\frac{1}{\sqrt{3}\cos {20}^o}=2k\cos {40}^o$, then $18k^4+162k^2+369$ is equal to

Answer 1

Given $\frac{1}{\sin {20}^o}+\frac{1}{\sqrt{3}\cos {20}^o}=2k\cos {40}^o$ ....(1)
Consider,$\frac{1}{\sin {20}^o}+\frac{1}{\sqrt{3}\cos {20}^o}$
$=\frac{\sqrt{3}\cos {20}^o+\sin {20}^o}{\sqrt{3}\sin {20}^o\cos {20}^o}$
$=\frac{\frac{\sqrt{3}}{2}\cos {20}^o+\frac{1}{2}\sin {20}^o}{\frac{\sqrt{3}}{4}\sin {40}^o}$
$=\frac{\sin {60}^o\cos {20}^o+\cos {60}^o\sin {20}^o}{\left(\sqrt{3}/4\right)\sin {40}^o}$
$=\frac{\sin {80}^0}{\frac{\sqrt{3}}{4}\sin {40}^0}$
$=\frac{8}{\sqrt{3}}\cos {40}^o$
So, by (1), we get
$k=\frac{4}{\sqrt{3}}$
$\Rightarrow 3k^2=16$
So $18k^4+162k^2+369=1745$.