Given secθ = 13/12, calculate all other trigonometric ratios.

Let △ABC be a right-angled triangle (right-angled at B)
secθ = AC/AB = 13/12
Let AC = 13x and AB = 12x
AC² = AB² + BC²
BC² = AC² - AB²
BC² = (13x)² - (12x)²
BC² = 169x² - 144x²
BC² = 25x²
BC = √25x² = 5x
sinθ = BC/AC = 5x/13x = 5/13
cosθ = AB/AC = 12/13
tanθ = BC/AB = 5x/12x = 5/12
cotθ = 1/tanθ = 1/(5/12) = 12/5
cosec θ = AC/BC = 13x/5x = 13/5