If \(\sin\theta+\cos\theta=p\) and \(\sec\theta+\csc\theta=q\), prove that \(q(p^2-1)=2p\).
\(q(p^2-1)=2p\).
Step 1: We are given two equations:
Step 2: First, let us square the first equation:
\[(\sin\theta + \cos\theta)^2 = p^2\]
Expanding: \(\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = p^2\).
Step 3: Recall the identity: \(\sin^2\theta + \cos^2\theta = 1\).
So: \(1 + 2\sin\theta\cos\theta = p^2\).
Step 4: Rearranging, we get:
\(p^2 - 1 = 2\sin\theta\cos\theta\). (Equation A)
Step 5: Now, consider the second given expression:
\(q = \sec\theta + \csc\theta\).
Write \(\sec\theta = \dfrac{1}{\cos\theta}\), and \(\csc\theta = \dfrac{1}{\sin\theta}\).
So: \(q = \dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta}\).
Step 6: Take LCM:
\(q = \dfrac{\sin\theta + \cos\theta}{\sin\theta \cos\theta}\).
Step 7: But from Step 1, \(\sin\theta + \cos\theta = p\).
So: \(q = \dfrac{p}{\sin\theta \cos\theta}\). (Equation B)
Step 8: Now, multiply Equation (A) and (B):
\[ q(p^2 - 1) = \dfrac{p}{\sin\theta\cos\theta} \times (2\sin\theta\cos\theta) \]
Step 9: The \(\sin\theta\cos\theta\) cancels out:
\[ q(p^2 - 1) = 2p \]
Final Answer: Hence proved.