Question

If $x=a\sec$ $\theta$ + b tan $\theta$ and $y=b\sec$ $\theta$ + a tan $\theta$, then $x^2-y^2$ is equal to

• A. $4ab\sec \theta \tan \theta$
• B. $a^2-b^2$
• C. $b^2-a^2$
• D. $a^2+b^2$

Answer 1

The correct option is B $a^2-b^2$

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

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

$=a^2{\sec }^2\theta +2ab\sec \theta \tan \theta +b^{2}ta{n}^2\theta -(b^2{\sec }^2\theta +2ab\sec \theta \tan \theta +a^2{\tan }^2\theta )$

$=(a^2-b^2){\sec }^2\theta +(b^2-a^2){\tan }^2\theta$

$=(a^2-b^2)({\sec }^2\theta -{\tan }^2\theta ).....[\because a^2-b^2=-(b^2-a^2)]$

$=a^2-b^2$ $......(\because {\sec }^2\theta -{\tan }^2\theta =1)$

Hence option ${}^{\prime }{B}^{\prime }$ is the answer.