Multiple Choice Question (MCQ)

Find the value of trigonometric expression

\dfrac{1}{\cos 290\degree } +\dfrac{1}{\sqrt{3} \sin 250\degree }

  1. ×

    \dfrac{\sqrt{3} }{4}

  2. \dfrac{4}{\sqrt{3} }

  3. ×

    \dfrac{\sqrt{3} }{2}

  4. ×

    \dfrac{2}{\sqrt{3} }

Collected in the board: Trigonometry

Steven Zheng posted 58 minutes ago

Answer

  1. \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

Steven Zheng posted 41 minutes ago

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