Question

Eliminate $\theta$ from the relations $a\sec \theta =1-b\tan \theta$ and $a^2{\sec }^2\theta =5+b^2{\tan }^2\theta$.

Answer 1

$Given:{\left(a\sec \theta \right)}^2={(1-b\tan \theta )}^2=1-2b\tan \theta +b^2{\tan }^2\theta ⟶\left(1\right)$

$\to a^2{\sec }^2\theta =5+b^2{\tan }^2\theta ⟶\left(II\right)$

$\left(I\right)$ - $\left(II\right)$ {Subtraction}

$\to 0=-4-2b\tan \theta$

$\to \tan \theta =\frac{-2}{b}\Rightarrow {\sec }^2\theta =1+{\tan }^2\theta =1+\frac{4}{b^2}$

$\to a^2{\sec }^2\theta =5+b^2{\tan }^2\theta \Rightarrow a^2(1+\frac{4}{b^2})=5+b^2\times \frac{4}{b^2}$

$\to a^2{b}^2+4a^2={ab}^2$