If \[z=\sec \,(y-ax)+\tan (y+ax),\] then \[\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}-{{a}^{2}}\frac{{{\partial }^{2}}z}{\partial {{y}^{2}}}=\] [EAMCET 2002]

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

1.	If \[z=\sec \,(y-ax)+\tan (y+ax),\] then \[\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}-{{a}^{2}}\frac{{{\partial }^{2}}z}{\partial {{y}^{2}}}=\] [EAMCET 2002]
A.	z
B.	2z
C.	0
D.	?z
Answer» D. ?z

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If \[z=\sec \,(y-ax)+\tan (y+ax),\] then \[\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}-{{a}^{2}}\frac{{{\partial }^{2}}z}{\partial {{y}^{2}}}=\] [EAMCET 2002]

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