If u=sin ^{-1}left(frac{x^2+y^2}{x+y}right) t | vedclass

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(B) Given $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$.This implies $\sin u = \frac{x^2+y^2}{x+y}$.Let $f(x, y) = \sin u = \frac{x^2+y^2}{x+y}$.Here

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$	an u$

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(B) Given $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}ight)$.This implies $\sin u = \frac{x^2+y^2}{x+y}$.Let $f(x, y) = \sin u = \frac{x^2+y^2}{x+y}$.Here,$f(x, y)$ is a homogeneous function of degree $n = 1$ because $f(tx, ty) = \frac{(tx)^2+(ty)^2}{tx+ty} = t \frac{x^2+y^2}{x+y} = t^1 f(x, y)$.According to Euler's theorem for homogeneous functions,$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f$.Substituting $f = \sin u$ and $n = 1$:$x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = 1 \cdot \sin u$.Using the chain rule,$\frac{\partial}{\partial x}(\sin u) = \cos u \frac{\partial u}{\partial x}$ and $\frac{\partial}{\partial y}(\sin u) = \cos u \frac{\partial u}{\partial y}$.So,$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = \sin u$.Dividing both sides by $\cos u$:$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{\sin u}{\cos u} = 	an u$.

Element	Co-factor
$A$. $-1$	$(1)$ $-2$
$B$. $1$	$(2)$ $32$
$C$. $3$	$(3)$ $4$
$D$. $6$	$(4)$ $6$
	$(5)$ $-6$

If $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to:

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Match the following elements of the matrix $A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \end{bmatrix}$ with their co-factors and choose the correct answer.
Element Co-factor
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