Given #tanx=sqrt3/3, cosx=-sqrt3/2# to find the remaining trigonometric function?

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Answer 1 · 2017-01-10T12:43:09

#sinx=-1/2#

#cotx==sqrt(3)#

#secx=-2/3sqrt(3)#

#cscx=-2#

Since

#tanx>0 and cosx<0#,

since

#tanx=sinx/cosx#,

it is sinx<0.

Then
#sinx=color(red)-sqrt(1-cos^2x)=-sqrt(1-(-sqrt(3)/2)^2)=-sqrt(1-3/4)=-1/2#

#cotx=1/tanx=3/sqrt(3)=sqrt(3)#

#secx=1/cosx=-2/sqrt(3)=-2/3sqrt(3)#

#cscx=1/sinx=-2#