A) \[ta{{n}^{-1}}x+ta{{n}^{-1}}y=ta{{n}^{-1}}c\]
B) \[ta{{n}^{-1}}x-ta{{n}^{-1}}y=ta{{n}^{-1}}c\]
C) \[-ta{{n}^{-1}}x+ta{{n}^{-1}}y=ta{{n}^{-1}}c\]
D) None of these
Correct Answer: A
Solution :
[a] \[\because (1+{{x}^{2}})\frac{dy}{dx}+(1+{{y}^{2}})=0\] \[\Rightarrow (1+{{x}^{2}})dy=-(1+{{y}^{2}})dx\] \[\Rightarrow \frac{dy}{1+{{y}^{2}}}=-\frac{dx}{1+{{x}^{2}}}\] Integrating/we have \[\int{\frac{dy}{1+{{y}^{2}}}=\int{-\frac{dx}{1+{{x}^{2}}}}}\] \[\Rightarrow ta{{n}^{-1}}y=-ta{{n}^{-1}}x+{{\tan }^{-1}}c\] Hence, option [a] is correct.
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