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?

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Answer 1

সঠিক উত্তর হচ্ছে: 1

ব্যাখ্যা:

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