A) \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=c\]
B) \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=c\]
C) \[{{\sec }^{-1}}x+\text{cose}{{\text{c}}^{-1}}x=c\]
D) None of these
Correct Answer: B
Solution :
Given equation is \[\int{\frac{dy}{\sqrt{1-{{y}^{2}}}}+\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}=0}}\] Integrating we get, \[{{\sin }^{-1}}y+{{\sin }^{-1}}x=c\].
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