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:
\(x = \cot \theta + \tan \theta\)
- - - - - (1)
\(y = \sec \theta - \cos \theta\)
- - - - - (2)
- - - - - (1)
\(y = \sec \theta - \cos \theta\)
- - - - - (2)
\(x^2y = \)
\((x^2 y)^{\frac{2}{3}}= \frac{1}{\cos^2 \theta}\)
\(xy^2 = \frac{\sin^3 \theta}{\cos^3 \theta}\)
\((x y^2)^{\frac{2}{3}} =\)
LHS \(= (x^2 y)^{\frac{2}{3}} - (x y^2)^{\frac{2}{3}}\)
\(= 1 = \) RHS
Hence, proved.
Answer variants:
\(= \frac{1}{\sin \theta \cos \theta}\)
\(\frac{\sin^2 \theta}{\cos^2 \theta}\)
\(= \frac{1}{\cos^3 \theta}\)
\(= \frac{\cos^2 \theta}{\cos^2 \theta}\)
\(= \frac{\sin^2 \theta}{\cos \theta}\)