Question

The value of $\frac{1}{\cos {290}^o}+\frac{1}{\sqrt{3}\sin {250}^o}$ is?

• A. $\frac{2\sqrt{3}}{3}$
• B. $\frac{4\sqrt{3}}{3}$
• C. $\sqrt{3}$
• D. None

Answer 1

The correct option is B $\frac{4\sqrt{3}}{3}$

Here, $\frac{1}{\cos {290}^o}+\frac{1}{\sqrt{3}\sin {250}^o}$

$=\frac{1}{\cos (270+20{)}^o}+\frac{1}{\sqrt{3}\sin (270-20{)}^o}$

as we know, $\cos (270+A)=\sin A$

& $\sin (270-B)=-\cos B$

So, $=\frac{1}{\sin 20}+\frac{1}{\sqrt{3}(-\cos 20)}$

$=\frac{-\sqrt{3}\cos 20+\sin 20}{-\sqrt{3}\cos 20\cos 20}$

$=\frac{-(\sqrt{3}\cos 20-\sin 20)}{-\sqrt{3}\sin 20\cos 20}$

$=\frac{\sqrt{3}\cos 20-\sin 20}{\sqrt{3}\sin 20\cos 20}$

(Multiply & Divide in Numerator & denominator by $2$ we get. )

$=\frac{2(\frac{\sqrt{3}}{2}\cos 20-\frac{1}{2}\sin 20)}{\frac{\sqrt{3}}{2}\left(2\sin 20\cos 20\right)}$

$=\frac{2(\sin 60\cos 20-\cos 60\sin 20)}{\frac{\sqrt{3}}{2}\left(\sin 40\right)}$

$=\frac{4}{\sqrt{3}}\frac{\sin (60-20)}{\sin \left(40\right)}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$

So, value is $4\sqrt{3}/3$