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    Topics || Problems

    Show that \(\tan \theta + \cot \theta = \sec \theta \csc \theta \)

    Solution

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