Question #978dc

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Question #978dc

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Answer 1 · 2016-08-24T06:05:52

Explanation:

Given that $(- \sqrt{3}, 1)$ $(-sqrt3,1)$ is the cartesian coordinate of the point on the terminal side of the angle $θ$ $theta$ .It is in the third quadrant.

Here $x = - \sqrt{3} and y = 1$ $x=-sqrt3 and y =1$

So lrngth of the terminal side
$r = \sqrt{x^{2} + y^{2}} = \sqrt{(- \sqrt{3}) + 1^{2}} = 2$ $r=sqrt(x^2+y^2)=sqrt((-sqrt3)+1^2)=2$

$Here in respect of θ$ $"Here in respect of "theta$
$x \to adjacent$ $x->"adjacent"$

$y \to opposite$ $y->"opposite"$

$r \to hypotenuse$ $r->"hypotenuse"$

$sin θ = \frac{y}{r} = \frac{1}{2}$ $sintheta=y/r=1/2$

$cos θ = \frac{x}{r} = - \frac{\sqrt{3}}{2}$ $costheta=x/r=-sqrt3/2$

$tan θ = \frac{y}{x} = \frac{1}{- \sqrt{3}} - \frac{1}{\sqrt{3}}$ $tantheta=y/x=1/(-sqrt3)-1/sqrt3$

$csc θ = \frac{r}{y} = \frac{2}{1} = 2$ $csctheta=r/y=2/1=2$

$sec θ = \frac{r}{x} = - \frac{2}{\sqrt{3}}$ $sectheta=r/x=-2/sqrt3$

$cot θ = \frac{x}{y} = - \frac{\sqrt{3}}{1}$ $cottheta=x/y=-sqrt3/1$

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Question #978dc

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