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

Given, value of $sec\theta =\frac{13}{12}$

Suppose $△\mathrm{ABC}$ be right-angled (at B).

So, from the trigonometry, 

$\sec \theta =\frac{\mathrm{hypotenuse }}{\mathrm{base}}=\frac{AC}{AB}$

Now, it is given that, 

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

On comparing we get,

Hypotenuse side, $\mathit{AC}=13$

Base side, $\mathit{AB}=12$

So, by the Pythagoras theorem, 

${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2$

${\mathrm{BC}}^2=A{\mathrm{C}}^2-{\mathrm{AB}}^2$
 

${\mathrm{BC}}^2=169-144$
 

${\mathrm{BC}}^2=25$
 

${\mathrm{BC}}^2=\sqrt{25}$
 

${\mathrm{BC}}^2=5$
 

Now, value of $\sin \theta$
 

$\sin \theta =\frac{\mathit{BC}}{\mathit{AC}}=\frac{5}{13}$
 
Also,

$\cos \theta =\frac{\mathit{AB}}{\mathit{AC}}=\frac{12}{13}$
 
And,

$\tan \theta =\frac{\mathit{BC}}{\mathit{AB}}=\frac{5}{12}$
 

As we know that,
 

$\cot \theta =\frac{1}{\tan \theta }$
 

So,
 

$\cot \theta =\frac{1}{\frac{5}{12}}=\frac{12}{5}$
 

Again, we know that,
 

$\mathrm{cosec}\theta =\frac{\mathit{AC}}{\mathit{BC}}=\frac{13}{5}$