Question

If $x=a\text{sec A}\text{cos B}$, $y=b\text{sec A}\text{sin B}$ and $z=c\text{tan A}$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=$

• A. 0
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is D

1

$x=a\text{sec A}\text{cos B}\Rightarrow \frac{x}{a}=\text{sec A}\text{cos B}$

$y=b\text{sec A}\text{sin B}\Rightarrow \frac{y}{b}=\text{sec A}\text{sin B}$

$z=c\text{tan A}\Rightarrow \frac{z}{c}=\text{tan A}$

Squaring each of the equations we get,

$\frac{x^2}{a^2}={\text{sec}}^{2}A{\text{cos}}^{2}B,\frac{y^2}{b^2}={\text{sec}}^{2}A{\text{sin}}^{2}B,\frac{z^2}{c^2}={\text{tan}}^{2}A$

$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}={\text{sec}}^{2}A{\text{cos}}^{2}B+{\text{sec}}^{2}A{\text{sin}}^{2}B-{\text{tan}}^{2}A$

$={\text{sec}}^{2}A({\text{cos}}^{2}B+{\text{sin}}^{2}B)-{\text{tan}}^{2}A$

$={\text{sec}}^{2}A-{\text{tan}}^{2}A\dots \dots ({\text{cos}}^{2}B+{\text{sin}}^{2}B=1)$

$=1\dots \dots ({\text{sec}}^{2}A-{\text{tan}}^{2}A=1)$