If sin x

We have:

$\sin x=\frac{3}{5}, \tan y=\frac{1}{2}$ and $\frac{\pi}{2}

Thus, $x$ is in the second quadrant and $y$ is in the third quadrant.

In the second quadrant, $\cos x$ and $\tan x$ are negative.

In the third quadrant, $\sec y$ is negative.

$\therefore \cos x=-\sqrt{1-\sin ^{2} x}=-\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{-4}{5}$

$\tan x=\frac{3 / 5}{-4 / 5}=\frac{-3}{4}$

And, $\sec y=-\sqrt{1+\tan ^{2} y}=-\sqrt{1+\left(\frac{1}{2}\right)^{2}}=\frac{-\sqrt{5}}{2}$

$\therefore 8 \tan x-\sqrt{5} \sec y=8 \times \frac{-3}{4}-\sqrt{5} \times \frac{-\sqrt{5}}{2}=-6+\frac{5}{2}=-\frac{7}{2}$