Question #50d29

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Answer 1 · 2016-06-30T15:29:47

If the question is as follows

$(cos(-60)-tan(135))/(tan315+cos660)$

$=(cos60-tan(2*90-45))/(tan(4*90-45)+cos(8*90-60))$

$=(cos60-(-tan45))/(-tan45+cos60)$

$=(1/2-(-1))/(-1+1/2)$

$=(3/2)/(1/2)=3/2xx2/1=3$

Otherwise
if the question is as follows

$cos(-60)-tan135/tan315+cos660$

$=cos60-(tan(2*90-45))/(tan(4*90-45))+cos(8*90-60)$

$=cos60-(-tan45)/(-tan45)+cos60$

$=1/2-1+1/2=0$