If \(a(\tan\theta + \cot\theta) = 1\) and \(\sin\theta + \cos\theta = b\), then prove that \(2a = b^2 - 1\), where \(0^\circ < \theta < 90^\circ\).

Q.If \(a(\tan\theta + \cot\theta) = 1\) and \(\sin\theta + \cos\theta = b\), then prove that \(2a = b^2 - 1\), where \(0^\circ < \theta < 90^\circ\).

Given: \(a(\tan\theta + \cot\theta) = 1\) Or, \(a\left(\cfrac{\sin\theta}{\cos\theta} + \cfrac{\cos\theta}{\sin\theta}\right) = 1\) Or, \(a\left(\cfrac{\sin^2\theta + \cos^2\theta}{\cos\theta \sin\theta}\right) = 1\) Or, \(a\left(\cfrac{1}{\cos\theta \sin\theta}\right) = 1\) ⇒ \(a = \sin\theta \cos\theta\) Now, \(2a = 2\sin\theta \cos\theta\) ⇒ \(2a + 1 = 2\sin\theta \cos\theta + 1\) ⇒ \(2a + 1 = 2\sin\theta \cos\theta + \sin^2\theta + \cos^2\theta\) ⇒ \(2a + 1 = (\sin\theta + \cos\theta)^2\) ⇒ \(2a + 1 = b^2\) ∴ \(2a = b^2 - 1\) (Proved)