Mark against the correct answer in the following:
Question:

Mark $(\sqrt{)}$ against the correct answer in the following:

The general solution of the $\operatorname{DEx} \sqrt{1+\mathrm{y}^{2}} \mathrm{dx}+\mathrm{y} \sqrt{1+\mathrm{x}^{2}} \mathrm{dy}=0$ is

A. $\sin ^{-1} x+\sin ^{-1} y=C$

B. $\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=C$

C. $\tan ^{-1} x+\tan ^{-1} y=\mathrm{C}$.

D. None of these

Solution:

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