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

Consider a right-angle triangle $\triangle ABC ,$ right-angled at point $B$.

$\sec \theta=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle \theta}$

$\frac{13}{12}=\frac{ AC }{ AB }$

If $AC$ is $13 k , AB$ will be $12 k,$ where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle A B C$, we obtain

$(A C)^{2}=(A B)^{2}+(B C)^{2}$

$(13 k)^{2}=(12 k)^{2}+(B C)^{2}$

$169 k^{2}=144 k^{2}+B C^{2}$

$25 k^{2}=B C^{2}$

$BC =5 k$

$\sin \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$

$\cos \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{12 k}{13 k}=\frac{12}{13}$

$\tan \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Side adjacent to } \angle \theta}=\frac{ BC }{ AB }=\frac{5 k}{12 k}=\frac{5}{12}$

$\cot \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Side opposite to } \angle \theta}=\frac{ AB }{ BC }=\frac{12 k}{5 k}=\frac{12}{5}$

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Side opposite to } \angle \theta}=\frac{ AC }{ BC }=\frac{13 k}{5 k }=\frac{13}{5}$