If $\sin{x}=t, \quad x\in(\frac{3\pi}{2},2\pi),$ what is $\tan{x}?$ So, according to the interval in which $x$ lies, the value of the sine is negative, and positive for the cosine. Using trig identity I get that
$$\cos{x}=\pm\sqrt{1-\sin^2{x}}=\pm\sqrt{1-t^2}.$$
Knowing the signs of sine and cosine, I pick $\sqrt{1-t^2}$. Computing $\tan{x}$ I get 
$$\tan{x}=\frac{\sin{x}}{\cos{x}}=\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}=\frac{t}{\sqrt{1-t^2}}.$$
The pick answers are
a) $\frac{\sqrt{t^2}}{\sqrt{1-t^2}};$
b) $-\frac{t}{\sqrt{1-t^2}};$
c) $-\frac{|t|}{\sqrt{1-t^2}};$
d) None of the above.
Why is c) correct?

You correctly found that 
$$\tan x = \frac{t}{\sqrt{1 - t^2}}$$
Observe that since $x \in (\frac{3\pi}{2}, 2\pi)$, $t = \sin x < 0$.  Hence, $t = -|t|$.  Thus, 
$$\tan x = \frac{t}{\sqrt{1 - t^2}} = -\frac{|t|}{\sqrt{1 - t^2}}$$
