If \(\sin \theta = \cfrac{p^2 - q^2}{p^2 + q^2}\), then show that \(\cot \theta = \cfrac{2pq}{p^2 - q^2}\) where \(p > q\) and \(0^\circ < \theta < 90^\circ\).

Q.If \(\sin \theta = \cfrac{p^2 - q^2}{p^2 + q^2}\), then show that \(\cot \theta = \cfrac{2pq}{p^2 - q^2}\) where \(p > q\) and \(0^\circ < \theta < 90^\circ\).

\(\sin \theta = \cfrac{p^2 - q^2}{p^2 + q^2}\) Or, \( \csc \theta = \cfrac{p^2 + q^2}{p^2 - q^2} \) Or, \( \csc^2 \theta = \left( \cfrac{p^2 + q^2}{p^2 - q^2} \right)^2 \) Or, \( \csc^2 \theta = \cfrac{(p^2 + q^2)^2}{(p^2 - q^2)^2} \) Or, \( \csc^2 \theta - 1 = \cfrac{(p^2 + q^2)^2}{(p^2 - q^2)^2} - 1 \)

[Subtracting 1 from both sides]

Or, \( \cot^2 \theta = \cfrac{(p^2 + q^2)^2 - (p^2 - q^2)^2}{(p^2 - q^2)^2} \) Or, \( \cot^2 \theta = \cfrac{4p^2q^2}{(p^2 - q^2)^2} \) Or, \( \cot \theta = \cfrac{2pq}{p^2 - q^2} \) (Proved)