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Answer 1 · 2017-02-23T14:45:00

Since $sin x = \frac{2}{3} \to + v e and cos x < 0 \to - v e$ $sinx =2/3->+ve and cosx <0->-ve$

the angle x should be II quadrant

So $cos x = - \sqrt{1 - {sin}^{2} x} = - \sqrt{1 - \frac{4}{9}} = - \frac{\sqrt{5}}{3}$ $cosx=-sqrt(1-sin^2x)=-sqrt(1-4/9)=-sqrt5/3$

$sec x = \frac{1}{cos x} = - \frac{3}{\sqrt{5}}$ $secx=1/cosx=-3/sqrt5$

$csc x = \frac{1}{sin x} = \frac{3}{2}$ $cscx =1/sinx=3/2$

$tan x = \frac{sin x}{cos x} = \frac{\frac{2}{3}}{- \frac{\sqrt{5}}{3}} = - \frac{2}{\sqrt{5}}$ $tanx=sinx/cosx=(2/3)/(-sqrt5/3)=-2/sqrt5$

$cot x = \frac{1}{tan x} = - \frac{\sqrt{5}}{2}$ $cotx=1/tanx=-sqrt5/2$