If u = sin^{-1}left(frac{x}{y}right) + tan^{- | vedclass

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(A) Given the function $u(x, y) = \sin^{-1}\left(\frac{x}{y}ight) + 	an^{-1}\left(\frac{y}{x}ight)$.We observe that $u(tx, ty) = \sin^{-1}\left(\

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$0$

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(A) Given the function $u(x, y) = \sin^{-1}\left(\frac{x}{y}ight) + 	an^{-1}\left(\frac{y}{x}ight)$.We observe that $u(tx, ty) = \sin^{-1}\left(\frac{tx}{ty}ight) + 	an^{-1}\left(\frac{ty}{tx}ight) = \sin^{-1}\left(\frac{x}{y}ight) + 	an^{-1}\left(\frac{y}{x}ight) = u(x, y)$.This implies that $u$ is a homogeneous function of degree $n = 0$.According to Euler's Theorem for homogeneous functions,if $u$ is a homogeneous function of degree $n$,then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n \cdot u$.Since $n = 0$,we have $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \cdot u = 0$.Therefore,the correct option is $A$.

If $u = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right)$,then the value of $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ is

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