1. **Problem Statement:** Given $$u = \cos^{-1} \left( \frac{x+y}{\sqrt{x} + \sqrt{y}} \right)$$ and $$v = \sin^{-1} \left( \frac{x^2 + y^2}{x + y} \right),$$ show that $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + \frac{1}{2} \cot u = 0$$ and $$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = \tan v.$$ 2. **Step 1: Define intermediate variables and use chain rule** Let $$z = \frac{x+y}{\sqrt{x} + \sqrt{y}}$$ so that $$u = \cos^{-1}(z).$$ By chain rule, $$\frac{\partial u}{\partial x} = -\frac{1}{\sqrt{1-z^2}} \frac{\partial z}{\partial x}, \quad \frac{\partial u}{\partial y} = -\frac{1}{\sqrt{1-z^2}} \frac{\partial z}{\partial y}.$$ 3. **Step 2: Compute partial derivatives of $z$** Recall $$z = \frac{x+y}{\sqrt{x} + \sqrt{y}}.$$ Calculate $$\frac{\partial z}{\partial x} = \frac{(1)(\sqrt{x} + \sqrt{y}) - (x+y) \frac{1}{2\sqrt{x}}}{(\sqrt{x} + \sqrt{y})^2} = \frac{\sqrt{x} + \sqrt{y} - \frac{x+y}{2\sqrt{x}}}{(\sqrt{x} + \sqrt{y})^2}.$$ Similarly, $$\frac{\partial z}{\partial y} = \frac{1 \cdot (\sqrt{x} + \sqrt{y}) - (x+y) \frac{1}{2\sqrt{y}}}{(\sqrt{x} + \sqrt{y})^2} = \frac{\sqrt{x} + \sqrt{y} - \frac{x+y}{2\sqrt{y}}}{(\sqrt{x} + \sqrt{y})^2}.$$ 4. **Step 3: Form the combination $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$** Substitute the derivatives: $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = -\frac{1}{\sqrt{1-z^2}} \left(x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}\right).$$ Calculate $$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{1}{(\sqrt{x} + \sqrt{y})^2} \left[x \left(\sqrt{x} + \sqrt{y} - \frac{x+y}{2\sqrt{x}}\right) + y \left(\sqrt{x} + \sqrt{y} - \frac{x+y}{2\sqrt{y}}\right)\right].$$ Simplify inside the bracket: $$= x(\sqrt{x} + \sqrt{y}) + y(\sqrt{x} + \sqrt{y}) - \frac{x(x+y)}{2\sqrt{x}} - \frac{y(x+y)}{2\sqrt{y}}.$$ Rewrite terms: $$= (x+y)(\sqrt{x} + \sqrt{y}) - \frac{x(x+y)}{2\sqrt{x}} - \frac{y(x+y)}{2\sqrt{y}}.$$ Factor out $(x+y)$: $$= (x+y) \left(\sqrt{x} + \sqrt{y} - \frac{x}{2\sqrt{x}} - \frac{y}{2\sqrt{y}}\right).$$ Note that $$\frac{x}{2\sqrt{x}} = \frac{\sqrt{x}}{2}, \quad \frac{y}{2\sqrt{y}} = \frac{\sqrt{y}}{2}.$$ So inside the parenthesis: $$\sqrt{x} + \sqrt{y} - \frac{\sqrt{x}}{2} - \frac{\sqrt{y}}{2} = \frac{\sqrt{x}}{2} + \frac{\sqrt{y}}{2} = \frac{\sqrt{x} + \sqrt{y}}{2}.$$ Therefore, $$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = (x+y) \frac{\sqrt{x} + \sqrt{y}}{2 (\sqrt{x} + \sqrt{y})^2} = \frac{x+y}{2 (\sqrt{x} + \sqrt{y})}.$$ 5. **Step 4: Substitute back into the expression for $u$** Recall $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = -\frac{1}{\sqrt{1-z^2}} \cdot \frac{x+y}{2 (\sqrt{x} + \sqrt{y})}.$$ Note that $$\cos u = z = \frac{x+y}{\sqrt{x} + \sqrt{y}},$$ so $$\sin u = \sqrt{1 - \cos^2 u} = \sqrt{1 - z^2}.$$ Rewrite $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = -\frac{1}{\sin u} \cdot \frac{x+y}{2 (\sqrt{x} + \sqrt{y})}.$$ But since $$z = \frac{x+y}{\sqrt{x} + \sqrt{y}},$$ we have $$\frac{x+y}{\sqrt{x} + \sqrt{y}} = \cos u,$$ thus $$\frac{x+y}{2 (\sqrt{x} + \sqrt{y})} = \frac{\cos u}{2}.$$ Therefore, $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = -\frac{\cos u}{2 \sin u} = -\frac{1}{2} \cot u.$$ Rearranging, $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + \frac{1}{2} \cot u = 0,$$ which proves the first identity. 6. **Step 5: Prove the second identity for $v$** Given $$v = \sin^{-1} \left( \frac{x^2 + y^2}{x + y} \right).$$ Let $$w = \frac{x^2 + y^2}{x + y}.$$ By chain rule, $$\frac{\partial v}{\partial x} = \frac{1}{\sqrt{1 - w^2}} \frac{\partial w}{\partial x}, \quad \frac{\partial v}{\partial y} = \frac{1}{\sqrt{1 - w^2}} \frac{\partial w}{\partial y}.$$ 7. **Step 6: Compute partial derivatives of $w$** Using quotient rule, $$\frac{\partial w}{\partial x} = \frac{(2x)(x+y) - (x^2 + y^2)(1)}{(x+y)^2} = \frac{2x(x+y) - (x^2 + y^2)}{(x+y)^2}.$$ Simplify numerator: $$2x^2 + 2xy - x^2 - y^2 = x^2 + 2xy - y^2.$$ Similarly, $$\frac{\partial w}{\partial y} = \frac{(2y)(x+y) - (x^2 + y^2)(1)}{(x+y)^2} = \frac{2y(x+y) - (x^2 + y^2)}{(x+y)^2} = \frac{2xy + 2y^2 - x^2 - y^2}{(x+y)^2} = \frac{2xy + y^2 - x^2}{(x+y)^2}.$$ 8. **Step 7: Form the combination $x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y}$** Substitute: $$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = \frac{1}{\sqrt{1 - w^2}} \left(x \frac{\partial w}{\partial x} + y \frac{\partial w}{\partial y}\right).$$ Calculate numerator: $$x \frac{\partial w}{\partial x} + y \frac{\partial w}{\partial y} = \frac{1}{(x+y)^2} \left[x (x^2 + 2xy - y^2) + y (2xy + y^2 - x^2)\right].$$ Expand: $$= \frac{1}{(x+y)^2} \left(x^3 + 2x^2 y - x y^2 + 2 x y^2 + y^3 - x^2 y\right).$$ Group like terms: $$= \frac{1}{(x+y)^2} \left(x^3 + (2x^2 y - x^2 y) + (-x y^2 + 2 x y^2) + y^3\right) = \frac{1}{(x+y)^2} (x^3 + x^2 y + x y^2 + y^3).$$ Factor: $$x^3 + x^2 y + x y^2 + y^3 = (x + y)(x^2 + y^2).$$ So, $$x \frac{\partial w}{\partial x} + y \frac{\partial w}{\partial y} = \frac{(x + y)(x^2 + y^2)}{(x + y)^2} = \frac{x^2 + y^2}{x + y} = w.$$ 9. **Step 8: Substitute back to get the final expression** Recall $$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = \frac{w}{\sqrt{1 - w^2}}.$$ Since $$v = \sin^{-1}(w),$$ we have $$\sin v = w, \quad \cos v = \sqrt{1 - w^2}.$$ Therefore, $$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = \frac{\sin v}{\cos v} = \tan v,$$ which proves the second identity. **Final answers:** $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + \frac{1}{2} \cot u = 0,$$ $$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = \tan v.$$