Question

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

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{4}{\sqrt{3}}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B

$\frac{4}{\sqrt{3}}$

Explanation for the correct option:

Find the required value.

Given:$\frac{1}{\cos {290}^{\circ }}+\frac{1}{\sqrt{3}\sin {250}^{\circ }}$

$\frac{1}{\cos {290}^{\circ }}+\frac{1}{\sqrt{3}\sin {250}^{\circ }}=\frac{1}{\cos ({360}^{\circ }-{70}^{\circ })}+\frac{1}{\sqrt{3}\sin \left({180}^{\circ }+{70}^{\circ }\right)}$

$$\begin{gathered} =\frac{1}{\cos {70}^{\circ }}-\frac{1}{\sqrt{3}\sin {70}^{\circ }}\mathbf{\left[}\mathbf{\because }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{360}\mathbf{°}\mathbf{–}\mathbf{\theta }\mathbf{}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{}\mathbf{\&}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{180}\mathbf{°}\mathbf{+}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{\right]} \\ =\frac{\sqrt{3}\sin {70}^{\circ }-\cos {70}^{\circ }}{\sqrt{3}\cos {70}^{\circ }\sin {70}^{\circ }} \\ =\frac{2\left({\displaystyle \frac{\sqrt{3}}{2}}\sin {70}^{\circ }-{\displaystyle \frac{1}{2}}\cos {70}^{\circ }\right)}{{\displaystyle \frac{\sqrt{3}}{2}}\times 2\cos {70}^{\circ }\sin {70}^{\circ }} \\ =\frac{2\left(\cos {30}^{\circ }\sin {70}^{\circ }-\sin {30}^{\circ }\cos {70}^{\circ }\right)}{\frac{\sqrt{3}}{2}\times \sin {140}^{\circ }} \\ =\frac{2\sin ({70}^{\circ }-{30}^{\circ })}{\frac{\sqrt{3}}{2}\times \sin {140}^{\circ }}\mathbf{\left[}\mathbf{\because }\mathbf{sin}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{sinAcosB}\mathbf{-}\mathbf{cosAsinB}\mathbf{,}\mathbf{sin}\mathbf{2}\mathbf{A}\mathbf{=}\mathbf{2}\mathbf{sinAcosA}\mathbf{\right]} \\ =\frac{2\sin {40}^{\circ }}{\frac{\sqrt{3}}{2}\times \sin {140}^{\circ }} \\ =\frac{4}{\sqrt{3}}\times \frac{\sin ({180}^{\circ }-{40}^{\circ })}{\sin {140}^{\circ }}\left[\because \sin (180-\beta )=\mathrm{sin\beta }\right] \\ =\frac{4}{\sqrt{3}}\times \frac{\sin {140}^{\circ }}{\sin {140}^{\circ }} \\ =\frac{4}{\sqrt{3}} \end{gathered}$$

Hence, option (B) is the correct answer.