A) \[\frac{y-1}{x+1}\]
B) \[\frac{y+1}{x-1}\]
C) \[\frac{x-1}{y-1}\]
D) \[\frac{x-1}{y+1}\]
Correct Answer: A
Solution :
Given, \[{{\sec }^{-1}}\left( \frac{1+x}{1-y} \right)=a\] \[\Rightarrow \] \[\frac{1+x}{1-y}=\sec a\] \[\Rightarrow \] \[1+x=(1-y)\sec a\] \[\Rightarrow \] \[y\,\sec \,a=\sec \,a-1-x\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-1}{\sec \,a}=\frac{-1}{\left( \frac{1+x}{1-y} \right)}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-(1-y)}{(1+x)}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1-y}{1+x}\]
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