The answer
(a) proven
(b) \(\theta = 76.7^{\circ}\) or \(\theta = 166.7^{\circ}\)
O-Level A-Math 2021 Paper 1 Question 10 · Verified worked solution by the Genius Plus Academy teaching team
The question
(a) Prove \(\dfrac{\sin\theta}{1 - \cos\theta} - \dfrac{1}{\sin\theta} = \cot\theta\). [4]
(b) Hence solve \(\dfrac{\sin 2\theta}{1 - \cos 2\theta} - \dfrac{1}{\sin 2\theta} = -2\) for \(0^{\circ} \leqslant \theta \leqslant 180^{\circ}\). [4]
(a) Combine over a common denominator \(\sin\theta(1 - \cos\theta)\): \[\frac{\sin\theta}{1 - \cos\theta} - \frac{1}{\sin\theta} = \frac{\sin^2\theta - (1 - \cos\theta)}{\sin\theta(1 - \cos\theta)}.\] Using \(\sin^2\theta = 1 - \cos^2\theta\), the numerator is \(1 - \cos^2\theta - 1 + \cos\theta = -\cos^2\theta + \cos\theta = \cos\theta(1 - \cos\theta)\). So \[\frac{\cos\theta(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} = \frac{\cos\theta}{\sin\theta} = \cot\theta. \quad\textbf{(proven)}\]
(b) Replacing \(\theta\) by \(2\theta\) in the identity, the left side equals \(\cot 2\theta\), so \(\cot 2\theta = -2 \Rightarrow \tan 2\theta = -\tfrac12\). The basic angle is \(\tan^{-1}\tfrac12 = 26.57^{\circ}\). Since \(\tan 2\theta < 0\) and \(0^{\circ} \leqslant 2\theta \leqslant 360^{\circ}\), \(2\theta\) is in the second or fourth quadrant: \(2\theta = 180^{\circ} - 26.57^{\circ} = 153.43^{\circ}\) or \(2\theta = 360^{\circ} - 26.57^{\circ} = 333.43^{\circ}\). Hence \(\theta = 76.7^{\circ}\) or \(\theta = 166.7^{\circ}\) (1 d.p.).
Answer: (a) proven
(b) \(\theta = 76.7^{\circ}\) or \(\theta = 166.7^{\circ}\)
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
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It is a prove a trig identity question from Trig functions, identities & equations, worth 8 marks: 4 + 4.
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