If $\tan y = \frac{2t}{1 - t^2}$ and $\sin x = \frac{2t}{1 + t^2},$ then $\frac{dy}{dx} = $
- A$\frac{2}{1 + t^2}$
- B$\frac{1}{1 + t^2}$
- C$1$
- D$2$
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