A) \[si{{n}^{-1}}x.{{\sin }^{-1}}y=C\]
B) \[si{{n}^{-1}}x=C\,si{{n}^{-1}}y\]
C) \[si{{n}^{-1}}x-si{{n}^{-1}}y=C\]
D) \[{{\sin }^{-1}}x+si{{n}^{-1}}y=C\]
Correct Answer: D
Solution :
[d] We have \[\frac{dy}{dx}+\frac{\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}=0\] \[\frac{dy}{dx}=-\frac{\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\] \[\int{\frac{dy}{\sqrt{1-{{y}^{2}}}}}=-\int{\frac{dy}{\sqrt{1-{{x}^{2}}}}}\] \[si{{n}^{-1}}y=-si{{n}^{-1}}x+si{{n}^{-1}}C\] \[si{{n}^{-1}}x+si{{n}^{-1}}y=C\text{ }\left( Ans. \right)\] or answer may be written as \[si{{n}^{-1}}(x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}})=C'\] \[x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}}=C'\]
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