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The value of  \frac{1}{\cos 285^{0}}+\frac{1}{\sqrt{3}\sin 255^{0}}, is

  • Option 1)

    \sqrt{3}-\sqrt{2}

  • Option 2)

    2\sqrt{2}

  • Option 3)

    \frac{4\sqrt{2}}{\sqrt{3}}

  • Option 4)

    \frac{2\sqrt{2}}{3}

 

Answers (1)

As we learnt in 

Results from Compound Angles -

- wherein

These results can be obtained by compound angle formula i.e. sum and difference formulae.

 

 \frac{1}{\cos 285^{\circ}}+\frac{1}{\sqrt{3}\sin 255^{\circ}}

=\frac{1}{\cos 75^{\circ}}+\frac{1}{\sqrt{3}-\left ( \sin 75^{\circ} \right )}\, \, \, \, \, \, \begin{bmatrix} \because cos\ 285^{\circ}=+cos\ 75^{\circ} \\ sin\ 255^{\circ}=-sin\ 75^{\circ} \end{bmatrix} 

=\frac{1}{\cos 75^{\circ}}-\frac{1}{\sqrt{3}-\left ( \sin 75^{\circ} \right )}=\frac{1}{sin 15^{\circ}}-\frac{1}{\sqrt{3}\ sin 75^{\circ}}

=\frac{2\sqrt{2}}{\sqrt{3}-1}-\frac{1(2\sqrt{2})}{\sqrt3(\sqrt{3}+1)}

=2\sqrt{2}\left [ \frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}(\sqrt{3}+1)} \right ]

=\frac{2\sqrt{2}}{\sqrt{3}}\frac{\left [ (\sqrt{3}+1)\sqrt{3}-(\sqrt{3}-1) \right ]}{2}=\frac{4\sqrt{2}}{\sqrt{3}}


Option 1)

\sqrt{3}-\sqrt{2}

This option is incorrect

Option 2)

2\sqrt{2}

This option is incorrect

Option 3)

\frac{4\sqrt{2}}{\sqrt{3}}

This option is correct

Option 4)

\frac{2\sqrt{2}}{3}

This option is incorrect

Posted by

Vakul

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