UK
If ^@ x = r{\space} sec{\space} \alpha {\space}cos{\space}\beta^@, ^@y = r{\space} sec {\space}\alpha{\space} sin{\space}\beta^@ and ^@z = r{\space} tan {\space}\alpha^@, prove that
^@{\space}x^2 + y^2 - z^2 = r^2^@ .
Answer:
- Substituting the values of ^@x,^@ ^@y,^@ and ^@z^@ in ^@x^2 + y^2 - z^2 = r^2^@, we have ^@ \begin{aligned} x^2 + y^2 - z^2 &= r^2 {\space} sec^2 {\space}\alpha{\space} cos^2{\space} \beta + r^2 {\space} sec^2 {\space}\alpha{\space} sin^2 {\space}\beta - r^2 {\space} tan^2 {\space}\alpha\\ &= r^2 {\space} sec^2 {\space}\alpha{\space} (cos^2{\space} \beta + sin^2 {\space}\beta) - r^2 {\space} tan^2 {\space}\alpha\\ &= r^2 {\space} sec^2 {\space}\alpha - r^2 {\space} tan^2 {\space}\alpha &&[\because cos^2 {\space}\beta + sin^2{\space} \beta = 1] \\ &= r^2 {\space} ( sec^2 {\space}\alpha - tan^2 {\space}\alpha)\\ &= r^2 {\space} ( 1 ) &&[\because sec^2 {\space}\alpha - tan^2 {\space}\alpha = 1]\\ &= r^2 \end{aligned} ^@
- Hence, ^@ x^2 + y^2 - z^2 ^@ = ^@ r^2 ^@ .
