Let us assume a right angled triangle ABC, right angled at B

Given: 15 cot A = 8

Cot A = 8/15

cot A = Adjacent side/Opposite side

= AB/BC = 8/15

Let AB be 8k and BC will be 15k

Where, k is a positive real number.

According to the Pythagoras theorem,

AC²= AB² + BC²

Substitute the value of AB and BC

AC²= (8k)² + (15k)²

AC²= 64k² + 225k²

AC²= 289k²

AC = 17k

Now, we have to find the value of sin A and sec A

Sin (A) = Opposite side /Hypotenuse

Substitute the value of BC and AC, we get

Sin A = BC/AC = 15k/17k = 15/17

sin A = 15/17

Since secant or 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.

Sec (A) = Hypotenuse/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

AC/AB = 17k/8k = 17/8

sec (A) = 17/8