Given sec θ = 13/12 . Calculate all other trigonometric ratiosand please tell it asap

Dear Student

We know that
sec function is the reciprocal of the cos function
which is equal to the ratio of the length of the hypotenuse side to the adjacent side
Let us assume a right angled triangle ABC, right angled at B
sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

Let AC be 13k and AB will be 12k
Where, k is a positive real number.
According to the Pythagoras theorem,

the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle
and we get,

AC^2=AB^2+ BC^2
Substitute the value of AB and AC (13k)^2= (12k)^2+ BC^2
169k^2= 144k^2+ BC^2
169k^2= 144k^2+ BC^2
BC^2 =169k^2- 144k^2
BC^2= 25k^2
Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios So,
Sin θ =Opposite Side/Hypotenuse = BC/AC = 5/13
Cos θ =Adjacent Side/Hypotenuse = AB/AC = 12/13

tan θ =Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec θ =Hypotenuse/Opposite Side = AC/BC = 13/5

cotθ =Adjacent Side/Opposite Side = AB/BC = 12/5

these are required values

Thanks