Given #sinx=sqrt3/2, cosx=-1/2#, how do you find the remaining trigonometric functions?

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Answer 1 · 2017-01-26T20:00:54

#tanx=-sqrt3#

#secx=-2#

#cscx=2/3sqrt3#

#cotx=-sqrt3/3#

Since #tantheta=sintheta/costheta#, you get

#tanx=(sqrt3/cancel2)/(-1/cancel2)=-sqrt3#

Since #sectheta=1/costheta#, you get

#secx=1/(-1/2)=-2#

Since #csctheta=1/sintheta#, you get

#cscx=1/(sqrt3/2)=2/sqrt3=2/3sqrt3#

Since #cottheta=1/tantheta#, you get

#cotx=1/-sqrt3=-sqrt3/3#