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 $= 1/2-(-1)(-1)+1/2=0$ ..