Throughout this question \(\theta\) is measured in degrees. Show that \(\tan^2\theta - \sin^2\theta \equiv \tan^2\theta\sin^2\theta\). (4) Hence solve, for ...

Part (a) establishes an identity, part (b) uses it to factorise an equation, and part (c) asks for an argument rather than a calculation. The instruction "Without carrying out any further solution" in part (c) means the marks are for reasoning from part (a), and any attempt to solve the equation from scratch misses the point.

(a) Showing that \(\tan^{2}\theta - \sin^{2}\theta \equiv \tan^{2}\theta\sin^{2}\theta\) [4 marks]

The right-hand side is a product, so work on the left and aim to factorise. The first move is to express the tangent in terms of sine and cosine, since that is the only way to combine two functions that are currently unrelated:

\[\tan^{2}\theta - \sin^{2}\theta = \frac{\sin^{2}\theta}{\cos^{2}\theta} - \sin^{2}\theta \qquad \textbf{M1}\]

Put both terms over the common denominator \(\cos^{2}\theta\) and take out the common factor \(\sin^{2}\theta\):

\[= \frac{\sin^{2}\theta - \sin^{2}\theta\cos^{2}\theta}{\cos^{2}\theta} = \frac{\sin^{2}\theta\left(1 - \cos^{2}\theta\right)}{\cos^{2}\theta} \qquad \textbf{M1}\]

Now the Pythagorean identity \(\sin^{2}\theta + \cos^{2}\theta \equiv 1\), rearranged as \(1 - \cos^{2}\theta \equiv \sin^{2}\theta\), replaces the bracket:

\[= \frac{\sin^{2}\theta}{\cos^{2}\theta} \times \sin^{2}\theta \qquad \textbf{M1}\] \[= \tan^{2}\theta\sin^{2}\theta \qquad \textbf{A1}\]

as required. Since the identity is printed, all four marks are for the derivation. The three M marks are for the three distinct ideas, converting the tangent, forming a single fraction and factorising, and applying the Pythagorean identity, so partial credit is available even if the work stalls; the A1 is for a complete and correct chain ending in the printed form. Two errors are common: writing \(\tan^{2}\theta = \dfrac{\sin^{2}\theta}{\cos^{2}\theta}\) but then failing to give the second term the same denominator, and cancelling \(\sin^{2}\theta\) from numerator and denominator, which is impossible since the denominator contains only \(\cos^{2}\theta\).

(b) Solving \(\tan^{2}\theta - \sin^{2}\theta = 3\sin^{2}\theta\) for \(0^{\circ} \le \theta \le 360^{\circ}\) [4 marks]

Replace the left-hand side by the product from part (a), then bring everything to one side and factorise rather than dividing:

\[\tan^{2}\theta\sin^{2}\theta = 3\sin^{2}\theta \quad \Longrightarrow \quad \sin^{2}\theta\left(\tan^{2}\theta - 3\right) = 0 \qquad \textbf{M1}\]

This is the mark the question is built around. Dividing both sides by \(\sin^{2}\theta\) is the wrong turn: \(\sin^{2}\theta\) may be zero, and dividing by it silently destroys a whole family of solutions. Factorising keeps both cases. The first factor gives

\[\sin\theta = 0: \quad \theta = 0^{\circ},\ 180^{\circ},\ 360^{\circ} \qquad \textbf{A1}\]

and all three lie in the closed interval, so all three count. The second factor gives

\[\tan^{2}\theta = 3 \quad \Longrightarrow \quad \tan\theta = \pm\sqrt{3} \qquad \textbf{M1}\]

Taking both square roots is essential. Since the tangent has period \(180^{\circ}\), the positive case gives \(60^{\circ}\) and \(240^{\circ}\), and the negative case gives \(120^{\circ}\) and \(300^{\circ}\):

\[\theta = 60^{\circ},\ 120^{\circ},\ 240^{\circ},\ 300^{\circ} \qquad \textbf{A1}\]

There are therefore seven solutions in all. Check the least obvious one: at \(\theta = 180^{\circ}\), \(\tan\theta = 0\) and \(\sin\theta = 0\), so the equation reads \(0 - 0 = 3 \times 0\), which is true. Check one of the others: at \(\theta = 120^{\circ}\), \(\tan^{2}\theta = 3\) and \(\sin^{2}\theta = \tfrac{3}{4}\), so the left side is \(3 \times \tfrac{3}{4} = \tfrac{9}{4}\) by part (a) and the right side is \(3 \times \tfrac{3}{4} = \tfrac{9}{4}\). Note that \(\theta = 90^{\circ}\) and \(\theta = 270^{\circ}\) are not solutions, because the tangent is undefined there.

(c) Explaining why only \(\sin\theta = 0\) satisfies \(\tan^{2}\theta - \sin^{2}\theta = -\sin^{2}\theta\tan^{2}\theta\) [2 marks]

By part (a) the left-hand side is identically \(\tan^{2}\theta\sin^{2}\theta\), so the equation becomes

\[\tan^{2}\theta\sin^{2}\theta = -\sin^{2}\theta\tan^{2}\theta \quad \Longrightarrow \quad 2\sin^{2}\theta\tan^{2}\theta = 0 \qquad \textbf{M1}\]

The M1 is for reaching this equation by substituting the identity, which is the "without further solution" route the question demands. A product is zero only when one of its factors is zero, so either \(\sin\theta = 0\) or \(\tan\theta = 0\). But \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) vanishes exactly when its numerator vanishes, so \(\tan\theta = 0\) if and only if \(\sin\theta = 0\). The two conditions collapse into one, and the solutions are precisely those with \(\sin\theta = 0\) A1.

The instructive point is why the sign change on the right-hand side is so destructive. In part (b) the equation was a product equal to a non-zero multiple of \(\sin^{2}\theta\), which left room for \(\tan^{2}\theta = 3\). Here moving the right-hand side across doubles the same product instead of cancelling it, so nothing survives except the values that make the product itself zero. Those are \(\theta = 0^{\circ},\ 180^{\circ},\ 360^{\circ}\), the same three found in part (b).

Examination takeaway. Never divide a trigonometric equation by a factor that can be zero; move everything to one side and factorise, so the zero case is preserved. When a part says "without further solution", the expected answer is an argument built on the identity just proved, and a zero product is the tool that turns it into a complete one.