If \(\cot \theta + \tan \theta = x\) and \(\sec \theta - \cos \theta = y\), then prove that \((x^2 y)^{\frac{2}{3}} - (x y^2)^{\frac{2}{3}} = 1\).
Proof:
\(x = \cot \theta + \tan \theta\)
\(x^2=\)
\(y = \sec \theta - \cos \theta\)
\(y^2=\)
\(xy^2=\)
\(x^2y=\)
\((x^2 y)^{\frac{2}{3}}=\)
\((x y^2)^{\frac{2}{3}} =\)
\((x^2 y)^{\frac{2}{3}} - (x y^2)^{\frac{2}{3}} = 1\).
Answer variants:
\(\frac{\sin^3 \theta}{\cos^3 \theta}\)
\(\frac{1}{\cos^2 \theta}\)
\(\frac{1}{\cos^3 \theta}\)
\(\frac{\sin^2 \theta}{\cos \theta}\)
\(\frac{\sin^2 \theta}{\cos^2 \theta}\)
\(\frac{1}{\sin \theta \cos \theta}\)