If $$\sin x + \cosec x = 2,$$ then $$\sin^{17} x + \cosec^{18} x$$ is equal to:

$$\sin x + \cosec x = 2$$

$$=$$>  $$\sin x+\frac{1}{\sin x}=2$$

$$=$$>  $$\sin^2x+1=2\sin x$$

$$=$$>  $$\sin^2x-2\sin x+1=0$$

$$=$$>  $$\left(\sin x-1\right)^2=0$$

$$=$$>  $$\sin x-1=0$$

$$=$$>  $$\sin x=1$$ 

$$=$$>  $$\frac{1}{\operatorname{cosec}x}=1$$

$$=$$>  $$\operatorname{cosec}x=1$$

$$\therefore\ \sin^{17}x+\operatorname{cosec}^{18}x=\left(1\right)^{17}+\left(1\right)^{18}=1+1=2$$

Hence, the correct answer is Option C