MCQ
If $z = \sec \,(y - ax) + \tan (y + ax),$ then ${{{\partial ^2}z} \over {\partial {x^2}}} - {a^2}{{{\partial ^2}z} \over {\partial {y^2}}} = $
- A$z$
- B$2z$
- ✓$0$
- D$-z$
$\frac{{{\partial ^2}z}}{{\partial {x^2}}} = {a^2}{\sec ^3}(y - ax) + {a^2}\sec (y - ax){\tan ^2}(y - ax)$
$ + 2{a^2}{\sec ^2}(y + ax)\tan (y + ax)$
$\frac{{\partial z}}{{\partial y}} = \sec (y - ax)\tan (y - ax) + {\sec ^2}(y + ax)$
$\frac{{{\partial ^2}z}}{{\partial {y^2}}} = {\sec ^3}(y - ax) + \sec (y - ax){\tan ^2}(y - ax)$
$ + 2{\sec ^2}(y + ax)\tan (y + ax)$
$\therefore $ $\frac{{{\partial ^2}z}}{{\partial {x^2}}} - {a^2}\frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0$
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