USA
If ^@x^@ = ^@a{\space} cosec {\space} \theta {\space} sin{\space}\phi, ^@ ^@y^@ = ^@b{\space}cosec{\space}\theta{\space} cos{\space}\phi^@ and ^@z ^@ = ^@c{\space} cot{\space}\theta^@ then prove that ^@ \bigg( \dfrac { x^2 } { a^2 } + \dfrac { y^2 } { b^2 } \bigg) = \bigg( 1 + \dfrac { z^2 } { c^2 } \bigg).^@
Answer:
- We are given that @^ \begin{aligned} & \dfrac { x } { a } = cosec{\space}\theta{\space} sin{\space}\phi &&\ldots \text{(i)} \\ & \dfrac { y } { b } = cosec{\space}\theta{\space} cos{\space}\phi &&\ldots \text{(ii)} \\ & \dfrac { z } { c } = cot{\space}\theta &&\ldots \text{(iii)} \end{aligned} @^
- On squaring and adding ^@ \text{(i)} ^@ and ^@ \text{(ii)} ^@, we get @^ \begin{aligned} \bigg( \dfrac { x^2 } { a^2 } + \dfrac { y^2 } { b^2 } \bigg) & = cosec^2{\space}\theta(sin^2{\space}\phi + cos^2{\space}\phi) \\ & = cosec^2{\space}\theta &&[ \because sin^2{\space}\phi + cos^2{\space}\phi = 1 ]\\ & = (1 + cot^2{\space}\theta) && [\because cosec^2 = 1 + cot^2{\space}\theta]\\ & = \bigg( 1 + \dfrac { z^2 } { c^2 } \bigg) &&[\text{By equation (iii)}] \end{aligned} @^
- Hence,^@ \bigg( \dfrac { x^2 } { a^2 } + \dfrac { y^2 } { b^2 } \bigg) = \bf \bigg( 1 + \dfrac { z^2 } { c^2 } \bigg)^@.
