If \(sin \ \theta + cos \ \theta = p\) and \(sec \ \theta + cosec \ \theta = q\), then show that \(q(p^2 - 1) = 2p\).
Proof:
\(p=sin \ \theta + cos \ \theta\)
\(p^2=\) ---- (\(1\))
\(q=\)---- (\(2\))
\(LHS\) \(= q(p^2 - 1)\)
From equations (1) and (2)
\(LHS=\)
\(= 2p\)
\(=\) \(RHS\)
Hence, we proved.
Answer variants:
\(\frac{p}{sin \ \theta \ cos \ \theta}\)
\(\frac{p}{sin \ \theta \ cos \ \theta}(2 sin \ \theta \ cos \ \theta)\)
\(1 + 2 sin \ \theta \ cos \ \theta\)