1.

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
Answer» D.            2


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