Consider a right-angled triangle, right-angled at $B.$

$\cot A=\frac{\text { Side adjacent to } \angle A }{\text { Side opposite to } \angle A }$

$=\frac{A B}{B C}$

It is given that,

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

$\frac{A B}{B C}=\frac{8}{15}$

Let $AB$ be $8 k$. Therefore, $BC$ will be $15 k ,$ where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle ABC ,$ we obtain

$AC ^{2}= AB ^{2}+ BC ^{2}$

$=(8 k)^{2}+(15 k)^{2}$

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

$=289 k^{2}$

$AC =17 k$

$\sin A=\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AC }$

$=\frac{15 k}{17 k}=\frac{15}{17}$

$\sec A=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A }$

$=\frac{ AC }{ AB }=\frac{17}{8}$