$x=u\ cosv,y=u \ sinv\ Prove\ That \dfrac{\partial(u,v)}{\partial(x,y)}. \dfrac{\partial(x,y)}{\partial(u,v)} =1$ $ \dfrac{\partial(u,v)}{\partial(x,y)} =\begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\dfrac{\partial v} {\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} \ \dfrac{\partial(x,y)}{\partial(u,v)} =\begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\dfrac{\partial y} {\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} \ $ $x^2+y^2= u^2(cos^2v+sin^2v)=u^2\ \therefore u=\sqrt{(x^2+y^2)}\ \dfrac{y}{x} =\dfrac{\ sinv}{\ cosv}= \ tanv\ v=tan^{-1}(\dfrac{y}{x})\ $ $u=\sqrt{x^2+y^2}, \dfrac{\partial u}{\partial x} = \dfrac{x}{\sqrt{x^2+y^2}}, \dfrac{\partial u}{\partial y} = \dfrac{y}{\sqrt{x^2+y^2}}, \ $ $ v=tan^{-1}(\dfrac{y}x),\dfrac{\partial v}{\partial x} =\dfrac{x^2}{x^2+y^2}(-\dfrac{y}{x^2})=- \dfrac{y}{x^2+y^2}\ $ $x =u\ cosv,\dfrac{\partial x}{\partial u} =\ cosv,\dfrac{\partial x}{\partial v} =-u \ sinv \$

$y=u\ sinv, \dfrac{\partial y}{\partial u} = \ sinv,\dfrac{\partial x}{\partial u} =u \ cosv \\ $

$\dfrac{\partial(u,v)}{\partial(x,y)}. \dfrac{\partial(x,y)}{\partial(u,v)} =\begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\\dfrac{\partial v} {\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} .\begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y} {\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} $

$=\begin{bmatrix}\dfrac{x}{\sqrt{x^2+y^2}}&\dfrac{y}{\sqrt{x^2+y^2}}\\ -\dfrac{y}{x^2+y^2}&\dfrac{x}{x^2+y^2}\end{bmatrix}. \begin{bmatrix}cosv&-usinv\\sinv&ucosv\end{bmatrix}$

$=[\dfrac{x^2+y^2}{(x^2+y^2)(\sqrt{x^2+y^2})}].[u(cos^2v+sin^2v]\\ =\dfrac{u}{\sqrt{x^2+y^2}}\\=1(\because\ u=\sqrt{x^2+y^2} )\\ Proved$