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.
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