If $tan x = - \frac{2}{9}$ $tanx=-2/9$ $sin x > 0$ $sinx>0$ , how do you find the sin, cos, sec, csc, and cot?

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If $tan x = - \frac{2}{9}$ $tanx=-2/9$ $sin x > 0$ $sinx>0$ , how do you find the sin, cos, sec, csc, and cot?

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Answer 1 · 2015-04-16T15:17:42

If $tan (x)$ $tan(x)$ is negative and $sin (x)$ $sin(x)$ is positive,
$x$ $x$ must be in Quadrant 2
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and evaluation is only a matter of reading the graph.

Assuming $x ε [0, 2 π]$ $x epsilon [0,2pi]$

$sin (x) = \frac{2}{\sqrt{85}}$ $sin(x) = 2/sqrt(85)$

#cos(x) = - 9/sqrt(85)

$csc (x) = \frac{\sqrt{85}}{2}$ $csc(x) = sqrt(85)/2$

$sec (x) = - \frac{\sqrt{85}}{9}$ $sec(x) = -sqrt(85)/9$

$cot (x) = - \frac{9}{2}$ $cot(x) = -9/2$

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If $tan x = - \frac{2}{9}$ $tanx=-2/9$ $sin x > 0$ $sinx>0$ , how do you find the sin, cos, sec, csc, and cot?

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