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

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

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

$cscx=1/sinx=-2$

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