Question

If $x=a(\tan \theta +\sec \theta {)}^2$, $y=b(\tan \theta -\sec \theta {)}^2$ then prove that $x^2{y}^2=a^2{b}^2$.

Answer 1

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

$\to x^2{y}^2=a^2{b}^2{[{\tan }^2\theta -{\sec }^2\theta ]}^4=a^2{b}^2$

Thus $x^2{y}^2=a^2{b}^2$