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

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(A) 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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(A) Given $z = \sec(y - ax) + 	an(y + ax)$.First,find the partial derivatives with respect to $x$:$\frac{\partial z}{\partial x} = \sec(y - ax) 	an(y - ax) \cdot (-a) + \sec^2(y + ax) \cdot (a) = -a \sec(y - ax) 	an(y - ax) + a \sec^2(y + ax)$.$\frac{\partial^2 z}{\partial x^2} = -a [\sec(y - ax) \sec^2(y - ax) \cdot (-a) + 	an(y - ax) \cdot \sec(y - ax) 	an(y - ax) \cdot (-a)] + a [2 \sec(y + ax) \cdot \sec(y + ax) 	an(y + ax) \cdot (a)]$$= a^2 [\sec^3(y - ax) + \sec(y - ax) 	an^2(y - ax)] + 2a^2 \sec^2(y + ax) 	an(y + ax)$.Now,find the partial derivatives with respect to $y$:$\frac{\partial z}{\partial y} = \sec(y - ax) 	an(y - ax) + \sec^2(y + ax)$.$\frac{\partial^2 z}{\partial y^2} = \sec(y - ax) \sec^2(y - ax) + 	an(y - ax) \cdot \sec(y - ax) 	an(y - ax) + 2 \sec(y + ax) \cdot \sec(y + ax) 	an(y + ax)$$= \sec^3(y - ax) + \sec(y - ax) 	an^2(y - ax) + 2 \sec^2(y + ax) 	an(y + ax)$.Multiplying by $a^2$:$a^2 \frac{\partial^2 z}{\partial y^2} = a^2 [\sec^3(y - ax) + \sec(y - ax) 	an^2(y - ax) + 2 \sec^2(y + ax) 	an(y + ax)]$.Comparing the two expressions,we see that $\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2} = 0$.

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

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