Given sec θ = 13/12, calculate all other trigonometric ratios.
Let ΔABC be a right-angled triangle, right-angled at B.
We know that sec θ = OP/OM = 13/12 (Given)
Let OP be 13k and OM will be 12k where k is a positive real number.
By Pythagoras theorem we get,
OP2 = OM2 + MP2
(13k)2 = (12k)2 + MP2
169k2 - 144k2 = MP2
MP2 = 25k2
MP = 5
Now,
sin θ = MP/OP = 5k/13k = 5/13
cos θ = OM/OP = 12k/13k = 12/13
tan θ = MP/OM = 5k/12k = 5/12
cot θ = OM/MP = 12k/5k = 12/5
cosec θ = OP/MP = 13k/5k = 13/5
