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  • Trigonometry Solutions

    Topics || Problems

    Show that: \(\frac{{1 + {{\tan }^2}\theta }}{{\csc \theta }} = \sec \theta \tan \theta \)

    Solutions

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