Answer:
\[R.H.S.=\frac{{{p}^{2}}-1}{{{p}^{2}}+1}\] \[=\frac{{{(sec\theta +tan\theta )}^{2}}-1}{{{(sec\theta +tan\theta )}^{2}}+1}\] \[=\frac{{{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta -1}{{{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta +1}\] \[[\text{By}\,\,{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab]\] \[=\frac{(se{{c}^{2}}\theta -1)+ta{{n}^{2}}\theta +2\sec \theta \tan \theta }{{{\sec }^{2}}\theta +(1+ta{{n}^{2}}\theta )+2sec\theta tan\theta }\] \[=\frac{{{\tan }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta }{{{\sec }^{2}}\theta +{{\sec }^{2}}\theta +2\sec \theta \tan \theta }\] \[\left[ \begin{align} & {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \\ & {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\ \end{align} \right]\] \[=\frac{2{{\tan }^{2}}\theta +2\sec \theta \tan \theta }{2{{\sec }^{2}}\theta +2\sec \theta \tan \theta }\] \[=\frac{2\tan \theta (tan\theta +sec\theta )}{2\sec \theta (sec\theta +tan\theta )}=\frac{\tan \theta }{\sec \theta }\] \[=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}\] \[=\sin \theta =\text{L}\text{.H}\text{.S}\text{.}\] Hence Proved.
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