Find the value of trigonometric expression
\dfrac{1}{\cos 290\degree } +\dfrac{1}{\sqrt{3} \sin 250\degree }
\dfrac{\sqrt{3} }{4}
\dfrac{4}{\sqrt{3} }
\dfrac{\sqrt{3} }{2}
\dfrac{2}{\sqrt{3} }
\dfrac{1}{\cos 290\degree } +\dfrac{1}{\sqrt{3} \sin 250\degree }
=\dfrac{1}{\cos(360\degree - 70\degree) } +\dfrac{1}{\sqrt{3} \sin(180\degree + 70\degree ) }
=\dfrac{1}{\cos 70\degree } -\dfrac{1}{\sqrt{3} \sin 70\degree }
=\dfrac{\sqrt{3} \sin 70\degree -\cos 70\degree }{\cos 70\degree \cdot \sqrt{3}\sin 70\degree }
=\dfrac{2\bigg( \dfrac{\sqrt{3}}{2} \sin 70\degree -\dfrac{1}{2} \cos 70\degree \bigg) }{2\sin 70\degree\cos 70\degree \cdot \dfrac{\sqrt{3}}{2} }
=\dfrac{4\bigg( \cos 30\degree \sin 70\degree -\sin 30\degree \cos 70\degree \bigg) }{\sin 140\degree \cdot \sqrt{3}}
=\dfrac{4\sin(70\degree -30\degree ) }{\sin(180\degree -40\degree ) \cdot \sqrt{3}}
=\dfrac{4\sin 40\degree }{\sin 40\degree \cdot \sqrt{3} }
=\dfrac{4}{\sqrt{3} }
Therefore
B is the correct option