What does $cos(arctan(-1))+sin(arc csc(-1))$ equal?

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Answer 1 · 2016-03-10T08:55:05

$cos(arctan(-1))+sin(arc csc(-1))=-1+-1/sqrt2$

$cos(arctan(-1))$ means $cosalpha$ of an angle $alpha$ , where $tanalpha=-1$ . $tanalpha=-1$ for $alpha=(3pi)/4$ or $alpha=(-pi)/4$ .

$cos((3pi)/4)=-1/sqrt2$ and $cos((-pi)/4)=1/sqrt2$ . Hence $cos(arc tan(-1))=+-1/sqrt2$

$sin(arc csc(-1))$ means $sinbeta$ of an angle $beta$ , where $cscbeta=-1$ .

$cscbeta=-1$ for $beta=(3pi)/2$ and $sin((3pi)/2)=-1$ . Hence $sin(arc csc(-1))=-1$

Hence, $cos(arctan(-1))+sin(arc csc(-1))=-1+-1/sqrt2$

What does cos(arctan(-1))+sin(arc csc(-1))cos(arctan(-1))+sin(arc csc(-1)) equal?