If \(cosec \ \theta + cot \ \theta = P\), then prove that \(cos \ \theta = \frac{P^2 - 1}{P^2 + 1}\).
Proof:
Consider \(cosec \ \theta + cot \ \theta = P\) ---- (\(1\))
We know the identity \(cosec^2 \ \theta - cot^2 \ \theta =\)
\((cosec \ \theta + cot \ \theta)(cosec \ \theta - cot \ \theta) =\)
\(P(cosec \ \theta - cot \ \theta) =\)
\(cosec \ \theta - cot \ \theta = \frac{1}{P}\) ---- (\(2\)) [Using equation (\(1\))]
Adding equations (\(1\)) and (\(2\)), we get:
\(cosec \ \theta = P + \frac{1}{P}\)
\( cosec \ \theta = \frac{P^2 + 1}{P}\)
\(cosec \ \theta = \frac{P^2 + 1}{2P}\) ---- (\(3\))
Subtracting equation (\(2\)) from (\(1\)), we get:
\(cot \ \theta = P - \frac{1}{P}\)
\(cot \ \theta = \frac{P^2 - 1}{P}\)
\(cot \ \theta = \frac{P^2 - 1}{2P}\) ---- (\(4\))
Dividing equation (\(4\)) by (\(3\)), we get:
\(\frac{cot \ \theta}{cosec \ \theta} = \frac{P^2 - 1}{2P} \times \frac{2P}{P^2 + 1}\)
\(\frac{\left(\frac{cos \ \theta}{sin \ \theta} \right)}{\left(\frac{1}{sin \ \theta} \right)} = \frac{P^2 - 1}{P^2 + 1}\)
\(cos \ \theta = \frac{P^2 - 1}{P^2 + 1}\)
Hence, we proved.