What does #-sin(arccos(2))+tan("arccsc"(4))# equal?

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Answer 1 · 2018-06-14T14:00:59

#-sin(arccos(2))+tan("arccsc"(4))=1/15sqrt15-sqrt3i#

We use the following identities to evaluate the above expression:

#sinx=sqrt(1-cos^2x)#

#cotx=sqrt(csc^2x-1)#

#tanx=1/cotx#

So

#-sin(arccos(2))+tan("arccsc"(4))#

#=-sqrt(1-cos^2(arccos2))+1/sqrt(csc^2("arccsc"4)-1)#

#=-sqrt(1-4)+1/sqrt(16-1)#

#-sqrt(-3)+1/sqrt15#

#=1/15sqrt15-sqrt3i#