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

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(A) Given $z = \sec(y-ax) + 	an(y+ax)$.First, find the partial derivatives with respect to $x$:$\frac{\partial z}{\partial x} = \sec(y-ax)	an(y-ax)(-a) + \sec^2(y+ax)(a) = -a\sec(y-ax)	an(y-ax) + a\sec^2(y+ax)$.$\frac{\partial^2 z}{\partial x^2} = -a[\sec(y-ax)\sec^2(y-ax)(-a) + 	an(y-ax)\sec(y-ax)	an(y-ax)(-a)] + a[2\sec(y+ax)\sec(y+ax)	an(y+ax)(a)]$$= a^2\sec^3(y-ax) + a^2\sec(y-ax)	an^2(y-ax) + 2a^2\sec^2(y+ax)	an(y+ax)$.Next, find the partial derivatives with respect to $y$:$\frac{\partial z}{\partial y} = \sec(y-ax)	an(y-ax) + \sec^2(y+ax)$.$\frac{\partial^2 z}{\partial y^2} = \sec(y-ax)\sec^2(y-ax) + 	an(y-ax)\sec(y-ax)	an(y-ax) + 2\sec(y+ax)\sec(y+ax)	an(y+ax)$$= \sec^3(y-ax) + \sec(y-ax)	an^2(y-ax) + 2\sec^2(y+ax)	an(y+ax)$.Now, 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 $\frac{\partial^2 z}{\partial x^2}-a^2 \frac{\partial^2 z}{\partial y^2}$ is equal to

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