Given 15 cot A = 8, find sin A and sec A.

Let we have a triangle, 

$Cot A=\frac{8}{15}$

Since $Cot \left(A\right) =\frac{Adjacent side}{Opposite side}=\frac{AB}{BC}=\frac{8}{15}$

Let AB be 8k and BC will be 15k

where k is a positive real number.

Using Pythagoras theorem, we get

$A{C}^2=A{B}^2 + B{C}^2$

${\left(AC\right)}^2={\left(8k\right)}^2 + {\left(15k\right)}^2$

$A{C}^2=64k^2+225k^2$

$A{C}^2=289k^2$

Thus, $AC =17k$

$Sin \left(A\right) =\frac{Opposite side}{Hypotenuse}= \frac{BC}{AC}=\frac{15k}{17k}=\frac{15}{17}$

$Sec \left(A\right) =\frac{Hypotrnuse}{Adjacent side}=\frac{AC}{AB}=\frac{17k}{8k}=\frac{17}{8}$