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

Question

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

Aicharya · Accepted Answer

We have, $ \begin{aligned} & z=\sec (y-a x)+	an (y+a x) \ & \frac{\partial z}{\partial x}=-a \sec (y-a x) 	an (y-a x)+a \sec ^2(y+a x) \ & \frac{\partial^2 z}{\partial x^2}=-a\left[-a \sec (y-a x) 	an ^2(y-a x)ight. \ & \left.\sec ^2(y+a x) 	an (y+a x) \quad-a \sec ^3(y-a x)ight]+2 a^2 \ & \operatorname{Similarly}^2 \ & \frac{\partial^2 z}{\partial y^2}=\sec (y-a x) 	an ^2(y-a x) \ & \quad+\sec ^3(y-a x)+2 \sec ^2(y+a x) 	an (y+a x) \ & 	herefore \quad \frac{\partial^2 z}{\partial x^2}-a^2 \frac{\partial^2 z}{\partial y^2}=0 \end{aligned} $

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

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