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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
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