question_answer

If \[sec\theta +tan\theta =p\], prove that \[\sin \theta =\frac{{{p}^{2}}-1}{{{p}^{2}}+1}\]

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.