If \(\cot \theta + \tan \theta = y\) and \(\sec \theta - \cos \theta = z\), then check that \((y^2 z)^{\frac{2}{3}} - (y z^2)^{\frac{2}{3}} = 1\).
Proof:
\(y = \cot \theta + \tan \theta\)
Simplifying, we get:
\(y =\)
\(z = \sec \theta - \cos \theta\)
\(z =\)
\(y^2z=\)
\(yz^2=\)
\((y^2 z)^{\frac{2}{3}}=\)
\((y z^2)^{\frac{2}{3}} =\)
\((y^2 z)^{\frac{2}{3}} - (y z^2)^{\frac{2}{3}} = 1\)
Thus, LHS \(=\) RHS.
Hence, we proved.
Answer variants:
\(\frac{1}{\cos^2 \theta}\)
\(\frac{\sin^2 \theta}{\cos \theta}\)
\(\frac{\sin^3 \theta}{\cos^3 \theta}\)
\(\frac{1}{\cos^3 \theta}\)
\(\frac{\sin^2 \theta}{\cos^2 \theta}\)
\(\frac{1}{\sin \theta \cos \theta}\)