cot x = $\frac{3}{4}$ gives tan = $\frac{4}{3}$ $\csc^{2}$ x = 1 + $\cot^{2} x$ ⇒ $\csc^{2}$ x = 1 + $\frac{9}{16}$ = $\frac{25}{16}$ ⇒ cosec x = ± $\frac{5}{4}$ ⇒ cosec x = - $\frac{5}{4}$ [since x lies in third quadrant] cosec x = $-\frac{5}{4}$ gives sin x = $-\frac{4}{5}$ , cos x = cot x × sin x ⇒ cos x = $\frac{3}{4} \times \left(-\frac{4}{5}\right) = -\frac{3}{5}$ ⇒ sec x = $-\frac{5}{3}$