If z = sec(y - ax) + tan(y + ax),then frac{pa | vedclass

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(C) Given $z = \sec(y - ax) + 	an(y + ax)$.First,find the partial derivatives with respect to $x$:$\frac{\partial z}{\partial x} = \sec(y - ax) 	an(

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(C) Given $z = \sec(y - ax) + 	an(y + ax)$.First,find the partial derivatives with respect to $x$:$\frac{\partial z}{\partial x} = \sec(y - ax) 	an(y - ax) \cdot (-a) + \sec^2(y + ax) \cdot a = -a \sec(y - ax) 	an(y - ax) + a \sec^2(y + ax)$.Now,find the second partial derivative with respect to $x$:$\frac{\partial^2 z}{\partial x^2} = -a [\sec(y - ax) \sec^2(y - ax) \cdot (-a) + 	an(y - ax) \sec(y - ax) 	an(y - ax) \cdot (-a)] + a [2 \sec(y + ax) \cdot \sec(y + ax) 	an(y + ax) \cdot a]$$= a^2 \sec^3(y - ax) + a^2 \sec(y - ax) 	an^2(y - ax) + 2a^2 \sec^2(y + ax) 	an(y + ax)$.Next,find the partial derivatives with respect to $y$:$\frac{\partial z}{\partial y} = \sec(y - ax) 	an(y - ax) + \sec^2(y + ax)$.Now,find the second partial derivative with respect to $y$:$\frac{\partial^2 z}{\partial y^2} = \sec(y - ax) \sec^2(y - ax) + 	an(y - ax) \sec(y - ax) 	an(y - ax) + 2 \sec(y + ax) \cdot \sec(y + ax) 	an(y + ax)$$= \sec^3(y - ax) + \sec(y - ax) 	an^2(y - ax) + 2 \sec^2(y + ax) 	an(y + ax)$.Finally,calculate $\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2}$:$= [a^2 \sec^3(y - ax) + a^2 \sec(y - ax) 	an^2(y - ax) + 2a^2 \sec^2(y + ax) 	an(y + ax)] - a^2 [\sec^3(y - ax) + \sec(y - ax) 	an^2(y - ax) + 2 \sec^2(y + ax) 	an(y + ax)] = 0$.

If $z = \sec(y - ax) + \tan(y + ax)$,then $\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2} = $

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