question_answer

If \[x=p\text{ }sec\theta +q\text{ }tan\theta \] and \[y=p\text{ }tan\theta +q\text{ }sec\theta \] then prove that \[{{x}^{2}}-{{y}^{2}}={{p}^{2}}-{{q}^{2}}\].

Answer:

\[L.H.S.={{x}^{2}}-{{y}^{2}}\]

\[={{(p\,\,\sec \theta +q\,\,\tan \theta )}^{2}}-{{(p\,\,\tan \theta +q\,\,\sec \theta )}^{2}}\]

\[=({{p}^{2}}{{\sec }^{2}}\theta +{{q}^{2}}{{\tan }^{2}}\theta +2pq\,\,\sec \theta \,\,\tan \theta \,-\,({{p}^{2}}{{\tan }^{2}}\theta +{{q}^{2}}{{\sec }^{2}}\theta +2pq\,\,\sec \theta \,\,\tan \theta )\]

\[={{p}^{2}}{{\sec }^{2}}\theta +{{q}^{2}}{{\tan }^{2}}\theta +2pq\,\,\sec \theta \,\,\tan \theta \,-\,{{p}^{2}}{{\tan }^{2}}\theta -{{q}^{2}}{{\sec }^{2}}\theta -2pq\,\,\sec \theta \,\,\tan \theta \]

\[={{p}^{2}}({{\sec }^{2}}\theta -{{\tan }^{2}}\theta )-{{q}^{2}}({{\sec }^{2}}\theta -{{\tan }^{2}}\theta )\]

\[={{p}^{2}}-{{q}^{2}}\]                            \[[\because \,\,{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1]\]

= R.H.S.                                   Hence Proved.