Show that: \(1 + {\left( {\cot \theta } \right)^2} = {\left( {\csc \theta } \right)^2}\)
Solution:
\(\begin{array}{*{20}{l}}{1 + {{\cos }^2}\theta = {{\csc }^2}\theta }
\\\\{1 + \left( {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right) = {{\csc }^2}\theta }
\\\\{\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }} = {{\csc }^2}\theta }
\\\\\begin{array}{l}\frac{1}{{{{\sin }^2}\theta }} = {\csc ^2}\theta
\\\\{\csc ^2}\theta = {\csc ^2}\theta \end{array}\end{array}\)