Given $tan x = \frac{\sqrt{3}}{3}, cos x = - \frac{\sqrt{3}}{2}$ $tanx=sqrt3/3, cosx=-sqrt3/2$ to find the remaining trigonometric function?

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Given $tan x = \frac{\sqrt{3}}{3}, cos x = - \frac{\sqrt{3}}{2}$ $tanx=sqrt3/3, cosx=-sqrt3/2$ to find the remaining trigonometric function?

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Answer 1 · 2017-01-10T12:43:09

$sin x = - \frac{1}{2}$ $sinx=-1/2$

$cot x = = \sqrt{3}$ $cotx==sqrt(3)$

$sec x = - \frac{2}{3} \sqrt{3}$ $secx=-2/3sqrt(3)$

$csc x = - 2$ $cscx=-2$

Explanation:

Since

$tan x > 0 and cos x < 0$ $tanx>0 and cosx<0$ ,

since

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ ,

it is sinx<0.

Then
$sin x = - \sqrt{1 - {cos}^{2} x} = - \sqrt{1 - {(- \frac{\sqrt{3}}{2})}^{2}} = - \sqrt{1 - \frac{3}{4}} = - \frac{1}{2}$ $sinx=color(red)-sqrt(1-cos^2x)=-sqrt(1-(-sqrt(3)/2)^2)=-sqrt(1-3/4)=-1/2$

$cot x = \frac{1}{tan x} = \frac{3}{\sqrt{3}} = \sqrt{3}$ $cotx=1/tanx=3/sqrt(3)=sqrt(3)$

$sec x = \frac{1}{cos x} = - \frac{2}{\sqrt{3}} = - \frac{2}{3} \sqrt{3}$ $secx=1/cosx=-2/sqrt(3)=-2/3sqrt(3)$

$csc x = \frac{1}{sin x} = - 2$ $cscx=1/sinx=-2$

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Given $tan x = \frac{\sqrt{3}}{3}, cos x = - \frac{\sqrt{3}}{2}$ $tanx=sqrt3/3, cosx=-sqrt3/2$ to find the remaining trigonometric function?

1 Answer

Explanation:

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Given tanx=√33,cosx=−√32tanx=sqrt3/3, cosx=-sqrt3/2 to find the remaining trigonometric function?

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Given $tan x = \frac{\sqrt{3}}{3}, cos x = - \frac{\sqrt{3}}{2}$ $tanx=sqrt3/3, cosx=-sqrt3/2$ to find the remaining trigonometric function?