If $$\sin x - \cos x = 0, 0^\circ < x < 90^\circ$$ then the value of $$(\sec x + cosec  x)^2$$ is:

$$\sin x-\cos x=0$$

$$=$$>  $$\sin x=\cos x$$

$$=$$>  $$\frac{\sin x}{\cos x}=1$$

$$=$$>  $$\tan x=1$$

$$=$$>  $$x=45^{\circ\ }$$  ($$0^\circ < x < 90^\circ$$)

$$\therefore\ $$ $$\left(\sec x+\operatorname{cosec}x\right)^2$$ = $$\left(\sec45^{\circ\ }+\operatorname{cosec}45^{\circ\ }\right)^2$$

= $$\left(\sqrt{2}+\sqrt{2}\right)^2$$

= $$\left(2\sqrt{2}\right)^2$$

= $$4\left(2\right)$$

= $$8$$

Hence, the correct answer is Option C