Question

Find the value of cot(x) if sec(x)=-9/2 , sin(x)<0

Answer 1

$\sec x=-\dfrac92\implies\cos x=-\dfrac29$

$\sin^2x+\cos^2x=1\implies\sin x=\pm\sqrt{1-\cos^2x}$
$\sin x<0\implies\sin x=-\sqrt{1-\left(-\dfrac29\right)^2}=-\dfrac{\sqrt{77}}9$

$\cot x=\dfrac{\cos x}{\sin x}$
$\implies\cot x=\dfrac{-\frac29}{-\frac{\sqrt{77}}9}=\dfrac2{\sqrt{77}}$