If \(\cos x = \frac{1}{{\sqrt {1 + {t^2}} }}\) and \(\sin y = \frac{t}{{\sqrt {1 + {t^2}} }}\), then \(\frac{{dy}}{{dx}}\, = \)

1
- 1
2
\(\frac{{1 - t}}{{1 + {t^2}}}\)
3
\(\frac{1}{{1 + {t^2}}}\)
4
1

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