Introduction
This O-Level A-Maths question tests your ability to:
- recognise that \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\)
- cross-multiply correctly
- simplify and cancel terms safely
- solve for \(\theta\) within a given range
We follow the teacher’s exact working on the board.
The Question
Solve the equation:
\(\dfrac{\cos\theta + 4\sin\theta}{2\cos\theta + \sin\theta} = \cot\theta\)
for \(-\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2}\).
Step-by-Step Working (Teacher Working)
Start with:
\(\dfrac{\cos\theta + 4\sin\theta}{2\cos\theta + \sin\theta} = \dfrac{\cos\theta}{\sin\theta}\)
Cross-multiply:
\((\cos\theta + 4\sin\theta)\sin\theta = (2\cos\theta + \sin\theta)\cos\theta\)
Expand both sides:
\(\sin\theta\cos\theta + 4\sin^2\theta = 2\cos^2\theta + \sin\theta\cos\theta\)
Cancel \(\sin\theta\cos\theta\) on both sides:
\(4\sin^2\theta = 2\cos^2\theta\)
Divide both sides by \(2\cos^2\theta\):
\(\dfrac{4\sin^2\theta}{2\cos^2\theta} = 1 \quad \Rightarrow \quad 2\tan^2\theta = 1\)
So:
\(\tan^2\theta = \tfrac{1}{2}\)
Hence:
\(\tan\theta = \pm \dfrac{1}{\sqrt{2}}\)
Therefore:
\(\theta = \pm \tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)\)
Approximate value:
\(\tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) \approx 0.615\)
So within the given range:
\(\theta \approx -0.615,\; 0.615\)
Key Concepts
| Concept | Why It Matters |
|---|---|
| \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\) | Converts equation into algebra form |
| Cross-multiplication | Clears denominators safely |
| Cancelling common terms | Simplifies quickly |
| Solving \(\tan^2\theta = k\) | Gives two solutions: ± |
| Using the range | Confirms both answers are valid |
Tips for Students
- Convert \(\cot\theta\) before cross-multiplying
- Expand carefully and cancel only identical terms
- When you get \(\tan^2\theta\), remember you’ll get two values
- Always check the given range at the end
For Parents
Trig equation questions are common in O-Level A-Maths because they test both:
- algebra accuracy
- understanding of identities
- confidence with solutions in specific ranges
With enough practice, students learn to spot the pattern quickly and solve faster.