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If \(p = \sec\theta + \tan\theta\) and \(q = \text{cosec}\,\theta - \cot\theta\), then what is \((p - q - pq)\) equal to?

This question was previously asked in
CDS 2 2026 Maths Question Paper (13-Sep-2026)
The correct answer is

\(1\)

Write \(p = \dfrac{1+\sin\theta}{\cos\theta}\) and \(q = \dfrac{1-\cos\theta}{\sin\theta}\). Then \(p - q = \dfrac{\sin\theta(1+\sin\theta) - \cos\theta(1-\cos\theta)}{\sin\theta\cos\theta} = \dfrac{\sin\theta - \cos\theta + 1}{\sin\theta\cos\theta}\) (using \(\sin^2\theta+\cos^2\theta=1\)). Also \(pq = \dfrac{(1+\sin\theta)(1-\cos\theta)}{\sin\theta\cos\theta} = \dfrac{1-\cos\theta+\sin\theta-\sin\theta\cos\theta}{\sin\theta\cos\theta}\). So \(p - q - pq = \dfrac{(\sin\theta-\cos\theta+1)-(1-\cos\theta+\sin\theta-\sin\theta\cos\theta)}{\sin\theta\cos\theta} = \dfrac{\sin\theta\cos\theta}{\sin\theta\cos\theta} = 1\).

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Important Questions from Trigonometric Identities

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