A) \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=c\]
B) \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=c\]
C) \[{{\sec }^{-1}}x+\cos e{{c}^{-1}}x=c\]
D) none of the above
Correct Answer: B
Solution :
\[\frac{dy}{dx}+\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}=0\] \[\Rightarrow \] \[\frac{dy}{\sqrt{1-{{y}^{2}}}}+\frac{dx}{\sqrt{1-{{x}^{2}}}}=0\] On integrating, we get \[{{\sin }^{-1}}y+{{\sin }^{-1}}x=c\]
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