Given 15cotA=8, find sinA and secA.

Let ABC be a right angled triangle right angled at B. cotA=8/15
Let AB=8x and BC will be 15x.
AC^2=AB^2+BC^2
⇒AC^2=64x^2+225x^2
⇒AC=√289x^2=17x
⇒sinA=BC/AC=15x/17x=15/17
⇒secA=AC/AB=17x/8x=17/8