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

Let we have a triangle, 

$sec \theta =\frac{13}{12}$

Since,$Sec \left(A\right) =\frac{Hypotenuse}{Adjacent side}=\frac{AC}{AB}=\frac{13}{12}$

Let AB be 12k and AC will be 13k

where k is a positive real number.

Using Pythagoras theorem, we get

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

${\left(13k\right)}^2={\left(12\right)}^2 + {\left(BC\right)}^2$

$169k^2-144k^2 ={\left(BC\right)}^2$

${\left(BC\right)}^2=25k^2$

Thus, $BC =5k$

$Sin \theta =\frac{Opposite side}{Hypotenuse}= \frac{BC}{AC}=\frac{5k}{13k}=\frac{5}{13}$

$Cos \theta =\frac{Adjacent side}{Hypotenuse}= \frac{AB}{AC}= \frac{12k}{13k}=\frac{12}{13}$

$\tan \theta =\frac{Opposite side}{Adjacent side}=\frac{BC}{AB}=\frac{5k}{12k}=\frac{5}{12}$

$Cosec \theta =\frac{Hypotrnuse}{Opposite side}=\frac{AC}{BC}=\frac{13k}{5k}=\frac{13}{5}$

$Cot \theta =\frac{Adjacent side}{Opposite side}=\frac{AB}{BC}=\frac{12k}{5k}=\frac{12}{5}$