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

Let us draw a right angle triangle ABC, with ∠A = θ.

We know that,

sec θ = $\dfrac{\text{Hypotenuse}}{\text{Side adjacent to angle θ}}$

Substituting values we get,

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

Let AC = 13k and AB = 12k.

In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ (13k)² = (12k)² + BC²

⇒ BC² = 169k² - 144k²

⇒ BC² = 25k²

⇒ BC = $\sqrt{25k^2}$ = 5k.

We know that,

⇒ sin θ = $\dfrac{\text{Side opposite to angle θ}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{5k}{13k} = \dfrac{5}{13}$,

⇒ cos θ = $\dfrac{1}{\text{sec θ}} = \dfrac{1}{\dfrac{13}{12}} = \dfrac{12}{13}$

⇒ tan θ = $\dfrac{\text{Side opposite to angle θ}}{\text{Side adjacent to angle θ}} = \dfrac{BC}{AB} = \dfrac{5k}{12k} = \dfrac{5}{12}$.

⇒ cot θ = $\dfrac{1}{\text{tan θ}} = \dfrac{1}{\dfrac{5}{12}} = \dfrac{12}{5}$

⇒ cosec θ = $\dfrac{1}{\text{sin θ}} = \dfrac{1}{\dfrac{5}{13}} = \dfrac{13}{5}$.