Answers

2014-08-23T20:32:24+05:30
tan(60^o) =  \sqrt{3} \\  \\  tan^2(60^o) =  3 \\ \\  cos(30^o) =  \frac{ \sqrt{3} }{2}   \\ \\ cos^2(30^o) =  \frac{3}{4} \\ \\  sin(45^o) = cos(45^o) =  \frac{1}{ \sqrt{2} } \\ \\  sin^2(45^o) = cos^2(45^o) =  \frac{1}{2} \\ \\ sin(60^o) = \frac{ \sqrt{3} }{2} \\ \\ sin^2(30^o) = \frac{3}{4}

Now substitute the values

 \frac{ \frac{4}{(3)} +  \frac{1}{( \frac{3}{4} )} -  \frac{2}{ (\frac{1}{2} )} }{( \frac{3}{4} ) + ( \frac{1}{4} )} \\  \\  \frac{4}{3} + \frac{4}{3} - 4 \\  \\ \frac{8}{3} - 4 \\  \\ -\frac{4}{3}
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