Question

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x² − y²) = (a² − b²).

Answer 1

$\begin{array}{l}\\ \text{We have}{x}^2-y^2=[{\left(asec\theta +b\tan \theta \right)}^2-{\left(a\tan \theta +bsec\theta \right)}^2]\\ \text{= (}{a}^2{\sec }^2\theta +b^2{\tan }^2\theta +2ab\sec \theta \tan \theta )-(a^2{\tan }^2\theta +b^2{\sec }^2\theta +2ab\tan \theta \sec \theta )\\ \text{=}{a}^2{\sec }^2\theta +b^2{\tan }^2\theta -a^2{\tan }^2\theta -b^2{\sec }^2\theta \\ \text{=(}{a}^2{\sec }^2\theta -a^2{\tan }^2\theta )-(b^2{\sec }^2\theta -b^2{\tan }^2\theta )\\ =a^2({\sec }^2\theta -{\tan }^2\theta )-b^2({\sec }^2\theta -{\tan }^2\theta )\\ =a^2-b^2\text{[}\because {\sec }^2\theta -{\tan }^2\theta =1]\\ \text{Hence,}{x}^2-y^2=a^2-b^2\end{array}$