If \[\cos x=\frac{1}{\sqrt{1+{{t}^{2}}}}\]and \[\sin y=\frac{t}{\sqrt{1+{{t}^{2}}}}\], then \[\frac{dy}{dx}=\] [MP PET 1994]

If \[\cos x=\frac{1}{\sqrt{1+{{t}^{2}}}}\]and \[\sin y=\frac{t}{\sqrt{

1.	If \[\cos x=\frac{1}{\sqrt{1+{{t}^{2}}}}\]and \[\sin y=\frac{t}{\sqrt{1+{{t}^{2}}}}\], then \[\frac{dy}{dx}=\] [MP PET 1994]
A.	?1
B.	\[\frac{1-t}{1+{{t}^{2}}}\]
C.	\[\frac{1}{1+{{t}^{2}}}\]
D.	1
Answer» E.

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If \[\cos x=\frac{1}{\sqrt{1+{{t}^{2}}}}\]and \[\sin y=\frac{t}{\sqrt{1+{{t}^{2}}}}\], then \[\frac{dy}{dx}=\] [MP PET 1994]

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