$x=e^v \sec u $ $x_u = e^v \sec u \tan u $ $\ x_v = e^v \sec u $ $\ y= e^v \tan u $ $\ y_u= e^v \sec^2 u$ $ \ y_v = e^v\tan u$ $J\left ( \dfrac {x,y}{u,v} \right )= \dfrac {\partial (x,y)}{\partial (u,v)} \begin{vmatrix} \dfrac {\partial x}{\partial u} & \dfrac {\partial x}{\partial v} \ \dfrac {\partial y} {\partial u} & \dfrac{\partial y} {\partial v} \end{vmatrix} = \begin{vmatrix} x_u &x_v \y_u &y_v \end{vmatrix}= x_uy_v - x_v-y_u$ $= e^v \sec u \tan u \times e^v \tan u - e^v \sec u \ e^v \sec^2 u $ $\ =e^{2v}\sec u (\tan^2 u - \sec^2 u )= -e^{2v}\sec u$ $j\left ( \dfrac {u,v}{x,y} \right )= \dfrac {1}{j \left ( \frac {x,y}{u,v} \right ) } = \dfrac{1}{-e^{2v}\sec u}$ $\therefore , \ j\left ( \dfrac {u,v}{x,y} \right )= -e^{-2v}\cos u$