Show that: \(\frac{{1 + {{\tan }^2}\theta }}{{\csc \theta }} = \sec \theta \tan \theta \)
Solutions
\(\begin{array}{l}\frac{{1 + {{\tan }^2}\theta }}{{\csc \theta }} = \sec \theta \tan \theta
\\\\\frac{{1 + \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\frac{1}{{\sin \theta }}}} = \sec \theta \tan \theta
\\\\\frac{{\frac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\frac{1}{{\sin \theta }}}} = \sec \theta \tan \theta
\\\\\frac{{\frac{1}{{{{\cos }^2}\theta }}}}{{\frac{1}{{\sin \theta }}}} = \sec \theta \tan \theta
\\\\\frac{{\sin \theta }}{{{{\cos }^2}\theta }} = \sec \theta \tan \theta
\\\\\frac{{\sin \theta }}{{\cos \theta }}\left( {\frac{1}{{\cos \theta }}} \right) = \sec \theta \tan \theta
\\\\\sec \theta \tan \theta = \sec \theta \tan \theta \end{array}\)