If $$x = a{\text{ }}\sec \theta .\cos \phi ,$$     $$y = b{\text{ }}\sec \theta .sin\phi ,$$     $$z = c{\text{ tan}}\theta {\text{.}}$$   then the value of $$\frac{{{x^2}}}{{{a^2}}}$$  + $$\frac{{{y^2}}}{{{b^2}}}$$  - $$\frac{{{z^2}}}{{{c^2}}}$$  is?

Solution(By Examveda Team)

$$\eqalign{ & x = a{\text{ }}\sec \theta .\cos \phi \cr & y = b{\text{ }}\sec \theta .sin\phi \cr & z = c{\text{ tan}}\theta \cr & \frac{x}{a} = \sec \theta .\cos \phi \cr & \frac{y}{b} = \sec \theta .sin\phi \cr & \frac{z}{c} = {\text{tan}}\theta \cr & \therefore \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} \cr & \Rightarrow {\sec ^2}\theta .{\cos ^2}\phi + {\sec ^2}\theta .si{n^2}\phi - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta \left( {{{\cos }^2}\phi + si{n^2}\phi } \right) - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$