Question 260476
a){{{((cos120)/cos330) - ((sin90)/cos315) }}}
cos 120 = -1/2
cos 330 = sqrt(3)/2
sin 90 = 1
cos 315 = sqrt(2)/2
so, we get
{{{(-1/2)/(sqrt(3)/2) - 1/(sqrt(2)/2)}}}
which is
{{{-sqrt(3) - 2/sqrt(2)}}}
and finally
{{{-sqrt(3) - sqrt(2)}}}
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b) (sin^2 210)/ (3cos^2 120+cos150)
sin 210 = -1/2
sin^2(210) = 1/4
cos 120 = -1/2
cos^2(120) = 1/2
3cos^2(120) = 3/2
cos150 = -sqrt(3)/2
so, we get
{{{(1/4)/(3/2-sqrt(3)/2))}}}
which becomes 
{{{1/(2*(3-sqrt(3)))}}}
rationalizing, we get
{{{(3+ sqrt(3))/12}}}