Search any question & find its solution
Question:
Answered & Verified by Expert
If $z=\sec (y-a x)+\tan (y+a x)$, $\frac{\partial^2 z}{\partial x^2}-a^2 \frac{\partial^2 z}{\partial y^2}$ is equal to
Options:
Solution:
2057 Upvotes
Verified Answer
The correct answer is:
$0$
We have,
$$
\begin{aligned}
& z=\sec (y-a x)+\tan (y+a x) \\
& \frac{\partial z}{\partial x}=-a \sec (y-a x) \tan (y-a x)+a \sec ^2(y+a x) \\
& \frac{\partial^2 z}{\partial x^2}=-a\left[-a \sec (y-a x) \tan ^2(y-a x)\right. \\
& \left.\sec ^2(y+a x) \tan (y+a x) \quad-a \sec ^3(y-a x)\right]+2 a^2 \\
& \operatorname{Similarly}^2 \\
& \frac{\partial^2 z}{\partial y^2}=\sec (y-a x) \tan ^2(y-a x) \\
& \quad+\sec ^3(y-a x)+2 \sec ^2(y+a x) \tan (y+a x) \\
& \therefore \quad \frac{\partial^2 z}{\partial x^2}-a^2 \frac{\partial^2 z}{\partial y^2}=0
\end{aligned}
$$
$$
\begin{aligned}
& z=\sec (y-a x)+\tan (y+a x) \\
& \frac{\partial z}{\partial x}=-a \sec (y-a x) \tan (y-a x)+a \sec ^2(y+a x) \\
& \frac{\partial^2 z}{\partial x^2}=-a\left[-a \sec (y-a x) \tan ^2(y-a x)\right. \\
& \left.\sec ^2(y+a x) \tan (y+a x) \quad-a \sec ^3(y-a x)\right]+2 a^2 \\
& \operatorname{Similarly}^2 \\
& \frac{\partial^2 z}{\partial y^2}=\sec (y-a x) \tan ^2(y-a x) \\
& \quad+\sec ^3(y-a x)+2 \sec ^2(y+a x) \tan (y+a x) \\
& \therefore \quad \frac{\partial^2 z}{\partial x^2}-a^2 \frac{\partial^2 z}{\partial y^2}=0
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.