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

Given that: sec 0 = 13/12
Let sec 0 = 13k/12k, where k is a real number.
In △ABC, by Pythagoras theorem, we have
BC² = AC² – AB²
= (13k)² – (12k)²
= 169k² 144k²
= 25k²
⇒ BC = √(25k²) = 5k
Hence, sin 0 = BC/AC = 5k/13k = 5/13
cos 0 = AB/AC = 12k/13k = 12/13
tan 0 = BC/AB = 5k/12k = 5/12
cose 0 = AC/BC = 13k/5k = 13/5 and cot 0 = AB/BC = 12k/5k = 12/5

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