Given sec θ = `13/12` , calculate all other trigonometric ratios.

#### Solution

Let ΔABC be a right-angled triangle, right angled at point B.

It is given that:

sec θ = hypotenuse / side adjacent to ∠θ = AC/AB = 13/12

Let AC = 13k and AB = 12k, where k is a positive integer.

Applying pythagoras theorem in Δ ABC, we obtain:

AC^{2} = AB^{2} + BC^{2}

BC^{2} = AC^{2} - AB^{2}

BC^{2} = (13k)^{2} - (12k)^{2}

BC^{2} = 169 k^{2} - 144 k^{2}

BC^{2} = 25k^{2}

BC = 5k

BC = 5k

sinθ = `("side opposite to" ∠θ)/("hypotenuse") = ("BC")/("AC") = 5/13`

cosθ = `("side adjacent to" ∠θ)/("hypotenuse") = ("AB")/("AC") = 12/13`

tanθ = `("side opposite to" ∠θ)/("side adjacent to" ∠θ) = "(BC)"/"(AB)" = 5/12`

cotθ = `("side adjacent to" ∠θ)/("side opposite to" ∠θ) = ("AB")/("BC") = 12/5`

cosecθ = `("hypotenuse")/("side opposite to" ∠θ) = ("AC")/("BC") = 13/5`