sec x = $\frac{13}{5}$ gives cos x = $\frac{5}{13}$ $\tan^2 x$ = $\sec^2$ x - 1 ⇒ $\tan^2 x$ = $\frac{169}{25}$ - 1 = $\frac{144}{25}$ ⇒ tan x = ± $\frac{12}{5}$ ⇒ tan x = $\frac{-12}{5}$ ⇒ cot x = $\frac{-5}{12}$ [since x lies in fourth quadrant] sin x = tan x × cos x ⇒ sin x = $\frac{-12}{5} \times \frac{5}{13}$ = $\frac{-12}{13}$ ⇒ cosec x = $\frac{-13}{12}$