Given $\sec \theta = \frac{13}{12}$,calculate all other trigonometric ratios.

(N/A) Consider a right-angled triangle $\triangle ABC$,right-angled at point $B$.
$\sec \theta = \frac{\text{Hypotenuse}}{\text{Side adjacent to } \angle \theta} = \frac{AC}{AB} = \frac{13}{12}$.
Let $AC = 13k$ and $AB = 12k$,where $k$ is a positive constant.
Applying the Pythagoras theorem in $\triangle ABC$:
$(AC)^2 = (AB)^2 + (BC)^2$
$(13k)^2 = (12k)^2 + (BC)^2$
$169k^2 = 144k^2 + (BC)^2$
$(BC)^2 = 25k^2$
$BC = 5k$
Now,calculating the other trigonometric ratios:
$\sin \theta = \frac{\text{Side opposite to } \angle \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}$
$\cos \theta = \frac{\text{Side adjacent to } \angle \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$
$\tan \theta = \frac{\text{Side opposite to } \angle \theta}{\text{Side adjacent to } \angle \theta} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$
$\cot \theta = \frac{\text{Side adjacent to } \angle \theta}{\text{Side opposite to } \angle \theta} = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5}$
$\operatorname{cosec} \theta = \frac{\text{Hypotenuse}}{\text{Side opposite to } \angle \theta} = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}$