Question

Find the Value of

$\frac{ \sin(300) \tan(330) \sec(420) }{ \tan(135) \sin(135) \sin(210) \sec(315) }$
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Answer 1

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$= \frac{ \sin(300) \tan(330) \sec(420) }{ \tan(135) \sin(135) \sin(210) \sec(315) } \\ \\ = \frac{ \sin(270 + 30) \tan(360 - 30) \sec(360 + 60) }{ \tan(90 + 45) \sin(90 + 45) \sin(180 + 30) \sec(270 + 45) } \\ \\ = \frac{ \cos(30) \tan(30) \sec(60) }{ \frac{ \cos(45) }{ \sin(45) } \cos(45) \sin(30) \frac{1}{ \sin(45) } } \\ \\ = \frac{ \frac{ \sqrt{3} }{2} \times \frac{1}{ \sqrt{3} } \times 2 }{1 \times \frac{1}{ \sqrt{2} } \times \frac{1}{2} \times \sqrt{2 } } \\ \\ \implies 2$

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Answer 2

SOLUTION⬆️

Refer to the attachment.

Hope it helps ☺️