Correct Answer: $\frac{25}{4}$ units

Solution :
Given:
$AD \perp BC, AC=26$ units, $CD=10$ units, $BC=42$ units, $\angle DAC=x$ and $\angle B=y$
In $\triangle ACD$,
$AC^2 = CD^2 + AD^2$
⇒ $26^2=10^2$ + $AD^2$
⇒ $AD^2$ = 676 – 100
$\therefore$ $AD$ = 24 units
Now, $BD = BC – CD$
⇒ $BD$ = 42 – 10 = 32 units
In $\triangle ADB$,
$AB^2 = AD^2 + BD^2$
⇒ $AB^2 $ = $24^2+32^2$
⇒ $AB$ = $\sqrt{1600}=40$ units
Now, $\frac{6}{\cos x}-\frac{5}{\cos y}+8 \tan y$
= $6\sec x - 5 \sec y + 8 \tan y$
= $6×\frac{AC}{AD} - 5×\frac{AB}{BD} + 8×\frac{AD}{BD}$
= $6×\frac{26}{24}- 5×\frac{40}{32} + 8×\frac{24}{32}$
= $\frac{13}{2}-\frac{25}{4}+6$
= $\frac{25}{4}$ units
Hence, the correct answer is $\frac{25}{4}$ units.