If $\cot\theta + \cos\theta = p$ and $\cot\theta - \cos\theta = q$, prove that $p^2 - q^2 = 4\sqrt{pq}$
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer
Given: $\cot\theta + \cos\theta = p$ and $\cot\theta - \cos\theta = q$
LHS:
$$p^2 - q^2 = (\cot\theta + \cos\theta)^2 - (\cot\theta - \cos\theta)^2$$
$$= 4\cot\theta\cos\theta \quad \text{[using } (a+b)^2-(a-b)^2 = 4ab\text{]}$$
RHS:
$$pq = (\cot\theta + \cos\theta)(\cot\theta - \cos\theta) = \cot^2\theta - \cos^2\theta$$
$$= \frac{\cos^2\theta}{\sin^2\theta} - \cos^2\theta = \cos^2\theta\left(\frac{1}{\sin^2\theta}-1\right) = \frac{\cos^2\theta\cdot\cos^2\theta}{\sin^2\theta} = \cot^2\theta\cos^2\theta$$
$$\therefore\ 4\sqrt{pq} = 4\sqrt{\cot^2\theta\cos^2\theta} = 4\cot\theta\cos\theta$$
Since LHS $= 4\cot\theta\cos\theta =$ RHS, hence proved. $\blacksquare$
Source: Chapter 8, Section 8.4 Trigonometric Identities
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Explanation
- The key trick is using the identity $(a+b)^2-(a-b)^2 = 4ab$ for the LHS, giving $4\cot\theta\cos\theta$ quickly.
- For the RHS, compute $pq$ using the difference of squares $(a+b)(a-b)=a^2-b^2$, then simplify $\cot^2\theta - \cos^2\theta$ by expressing $\cot\theta$ as $\frac{\cos\theta}{\sin\theta}$.
- The result $\cot^2\theta\cos^2\theta$ under the square root gives exactly $\cot\theta\cos\theta$ (taking positive value for acute angles), matching LHS.
- Examiners award marks for: correct expansion (1 mark), correct simplification of $\sqrt{pq}$ (1 mark), and completing the proof (1 mark).