Question

If $x=a\sec \theta +b\tan \theta ,y=a\tan \theta +b\sec \theta$, then

• A. x + y = a + b
• B. x - y = a - b
• C. $x^2+y^2=a^2+b^2$
• D. $x^2-y^2=a^2-b^2$

Answer 1

The correct option is D $x^2-y^2=a^2-b^2$

$x=a\sec 0+b\tan 0,y=a\tan 0+b\sec 0$
Now, $x^2=(a\sec 0+b\tan 0{)}^2$
$x^2=a^{2}se{c}^{2}0+b^2{\tan }^{2}0+2ab\sec 0\tan 0$
$y^2=(a\tan 0+b\sec 0{)}^2$
$y^2=a^2{\tan }^{2}0+b^2{\sec }^{2}0+2ab\sec 0\tan 0$
Now, $x^2-y^2=(a^{2}se{c}^{2}0+b^2{\tan }^{2}0+2ab\sec 0\tan 0)-(a^2{\tan }^{2}0+b^2{\sec }^{2}0+2ab\sec 0\tan 0)$
$x^2-y^2=a^2(se{c}^{2}0-ta{n}^{2}0)-b^2(se{c}^{2}0-ta{n}^{2}0)$
$x^2-y^2=a^2-b^2$ (Since, $se{c}^{2}0-ta{n}^{2}0=1$)