$\frac{\sin (- 660 °) \tan (1050 °) \sec (- 420 °)}{\cos (225 °) \cos e c (315 °) \cos (570 °)} =$

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

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

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

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

answer is C.

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Detailed Solution

\frac{\sin (- 660 °) \tan (1050 °) \sec (- 420 °)}{\cos (225 °) \cos e c (315 °) \cos (570 °)}

$= \frac{(-) \sin (360 ° + 300 °) \tan (3 \times 360 ° - 30 °) \sec (360 ° + 60 °)}{\cos (180 ° + 45 °) cosec (360 ° - 45 °) \cos (360 ° + 210 °)}$

$= \frac{(-) \sin 300 ° \times (-) \tan 30 ° \times \sec 60 °}{(-) \cos 45 ° \times (-) cosec 45 ° \times \cos 210 °}$

$= \frac{\sin 60 ° \times (-) \tan 30 ° \times \sec 60 °}{(-) \cos 45 ° \times (-) cosec 45 ° \times (-) \cos 30 °}$

$= \frac{\frac{\sqrt{3}}{2} \times (-) \frac{1}{\sqrt{3}} \times 2}{\frac{1}{\sqrt{2}} \times \sqrt{2} \times (-) \frac{\sqrt{3}}{2}}$

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

$= \frac{2}{\sqrt{3}}$

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