Question #09e0d

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Answer 1 · 2016-07-24T14:52:14

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#-60^o# is in Quadrant IV (Q IV)i; Cosine is positive.

So #cos (-60^o)=cos(60^o)=1/2#

#135^0=(180-45)^o# is in Q II; tan is negative.

So, #tan(135^o)=tan(180-45)^o=-tan 45^o=-1#

#315^o=(360-45)^o# is in Q IV; tangent is negative.

So, #tan(315^o)=tan(360-45)^o=-tan45^o=-1#

#660^0=2X320-60)^0# is in Q IV; cos is positive.

So, #cos660^0=cos(2X360-60)^o=cos60^o=1/2#.

Now, the given expression# = 1/2-(-1)(-1)+1/2=0#..