Cot^2x•sec^2x convert each of these into sine and cosine and simplify

$\sec x=\frac1{\cos x}$

$\therefore \cot^2x\cdot\sec^2x= \frac{\cos^2x}{\sin^2x}\frac{1}{\cos^2x}=\frac{1}{\sin^2x}$

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c) the x-coordinate of the center of the circle increases by 3