$\cot x=\frac{3}{4}$

$\tan x=\frac{1}{\cot x}=\frac{1}{\left(\frac{3}{4}\right)}=\frac{4}{3}$

$1+\tan ^{2} x=\sec ^{2} x$

$\Rightarrow 1+\left(\frac{4}{3}\right)^{2}=\sec ^{2} x$

$\Rightarrow 1+\frac{16}{9}=\sec ^{2} x$

$\Rightarrow \frac{25}{9}=\sec ^{2} x$

$\Rightarrow \sec x=\pm \frac{5}{3}$

since $x$ lies in the $3^{\text {rd }}$ quadrant, the value of sec $x$ will be negative.

$\therefore \sec x=-\frac{5}{3}$

$\cos x=\frac{1}{\sec x}=\frac{1}{\left(-\frac{5}{3}\right)}=-\frac{3}{5}$

$\tan x=\frac{\sin x}{\cos x}$

$\Rightarrow \frac{4}{3}=\frac{\sin x}{\left(\frac{-3}{5}\right)}$

$\Rightarrow \sin x=\left(\frac{4}{3}\right) \times\left(\frac{-3}{5}\right)=-\frac{4}{5}$

$\cos ec\, x=\frac{1}{\sin x}=-\frac{5}{4}$