Question

If $a+b\tan \theta =\sec \theta$ and $b-a\tan \theta =3\sec \theta$, then find the value of $a^2+b^2$

Answer 1

Given equations are $a+b\tan \theta =\sec \theta$ and $b-a\tan \theta =3\sec \theta$
Squaring and adding, we get
$a^2+b^2{\tan }^2\theta +2ab\tan \theta +b^2+a^2{\tan }^2\theta -2ab\tan \theta ={\sec }^2\theta +9{\sec }^2\theta \Rightarrow (a^2+b^2)+(a^2+b^2){\tan }^2\theta =10{\sec }^2\theta \Rightarrow (a^2+b^2){\sec }^2\theta =10{\sec }^2\theta \Rightarrow (a^2+b^2)=10$