sec8A - 1 /sec4A - 1 =tan 8A / tan 4A
$\displaystyle \sec(8\alpha)-\frac{1}{\sec(4\alpha)}-1=\frac{\tan(8\alpha)}{\tan(4\alpha)}$ is false in general; see the graphs. Now, I know you want to prove $\displaystyle \frac{\sec(8\alpha)-1}{\sec(4\alpha)-1}=\frac{\tan(8\alpha)}{\tan(4\alpha)}$, but how can you deal with trigonometry if you don't know the order of arithmetic operations? The second equality is also false in general; see the graphs.