Question

If $x=a\text{sec}\theta +b\text{tan}\theta$ and $y=a\text{tan}\theta +b\text{sec}\theta$, then $\frac{x^2-y^2}{a^2-b^2}=$ ___

• A. 2
• B. 1
• C. 0
• D. - 1

Answer 1

The correct option is B

1

$x=a\text{sec}\theta +b\text{tan}\theta ,y=a\text{tan}\theta +b\text{sec}\theta$

Squaring both sides we get,

$x^2=(a\text{sec}\theta +b\text{tan}\theta {)}^2,y^2=(a\text{tan}\theta +b\text{sec}\theta {)}^2$

$\begin{array}{cc}\therefore & x^2=a^2{\text{sec}}^2\theta +2ab\text{sec}\theta \text{tan}\theta +b^2{\text{tan}}^2\theta \\ & y^2=a^2{\text{tan}}^2\theta +2ab\text{sec}\theta \text{tan}\theta +b^2{\text{sec}}^2\theta \\ & \underset{–}{---}\\ \text{Subtracting},& x^2-y^2=a^2({\text{sec}}^2\theta -{\text{tan}}^2\theta )+b^2({\text{tan}}^2\theta -{\text{sec}}^2\theta )\end{array}$

$x^2-y^2=a^2({\text{sec}}^2\theta -{\text{tan}}^2\theta )-b^2{\text{(sec}}^2\theta -{\text{tan}}^2\theta )$

$x^2-y^2=a^2-b^2\dots \dots ({\text{sec}}^2\theta -{\text{tan}}^2\theta =1)$

$\therefore \frac{x^2-y^2}{a^2-b^2}=1$