A-Level Mathematics May/June 2025 Q8(a): Prove the identity cot²θ - tan²θ ≡ 4 cot 2θ cosec 2θ.

1 answer

• ▲0
  accepted ✓

  Proving the Identity

  To prove the identity $\cot^2\theta - \tan^2\theta \equiv 4 \cot 2\theta \operatorname{cosec} 2\theta$, we start from the left-hand side (LHS).

  $LHS = \cot^2\theta - \tan^2\theta$

  Express the terms using sine and cosine: $LHS \equiv \frac{\cos^2\theta}{\sin^2\theta} - \frac{\sin^2\theta}{\cos^2\theta}$

  Combine the two fractions by finding a common denominator, $\sin^2\theta\cos^2\theta$: $LHS \equiv \frac{\cos^4\theta - \sin^4\theta}{\sin^2\theta\cos^2\theta}$

  Factorise the numerator as a difference of two squares, $(\cos^2\theta)^2 - (\sin^2\theta)^2$: $LHS \equiv \frac{(\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)}{\sin^2\theta\cos^2\theta}$

  Now, apply standard trigonometric identities. In the numerator, use $\cos^2\theta - \sin^2\theta \equiv \cos 2\theta$ and $\cos^2\theta + \sin^2\theta \equiv 1$. In the denominator, use the double angle formula $\sin 2\theta \equiv 2\sin\theta\cos\theta$, which implies $\sin^2\theta\cos^2\theta \equiv (\frac{1}{2}\sin 2\theta)^2 \equiv \frac{1}{4}\sin^2 2\theta$.

  $LHS \equiv \frac{(\cos 2\theta)(1)}{\frac{1}{4}\sin^2 2\theta}$

  $LHS \equiv \frac{4\cos 2\theta}{\sin^2 2\theta}$

  Finally, split the expression to match the form of the right-hand side (RHS): $LHS \equiv 4 \left( \frac{\cos 2\theta}{\sin 2\theta} \right) \left( \frac{1}{\sin 2\theta} \right)$

  Using the definitions $\cot x \equiv \frac{\cos x}{\sin x}$ and $\operatorname{cosec} x \equiv \frac{1}{\sin x}$: $LHS \equiv 4 \cot 2\theta \operatorname{cosec} 2\theta$

  $LHS \equiv RHS$

  Thus, the identity is proven.

  ---

  How the marks are awarded

  • M1 — The first method mark is for correctly expressing the LHS in terms of sine and cosine, as shown in the first line of working: $... \equiv \frac{\cos^2\theta}{\sin^2\theta} - \frac{\sin^2\theta}{\cos^2\theta}$.
  • M1 — The second method mark is for combining the fractions over a common denominator and correctly factorising the numerator using the difference of two squares, leading to the expression $\frac{(\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)}{\sin^2\theta\cos^2\theta}$.
  • M1 — The third method mark is for correctly applying double angle formulae to both the numerator and denominator to get an expression entirely in terms of $2\theta$. This involves using $\cos 2\theta$ and $\sin^2 2\theta$, leading to the intermediate step $\frac{4\cos 2\theta}{\sin^2 2\theta}$.
  • A1 — The final accuracy mark is for correctly manipulating the expression from the previous step into the required form of the RHS, by splitting the fraction into $\frac{\cos 2\theta}{\sin 2\theta}$ and $\frac{1}{\sin 2\theta}$, and concluding the proof.

  Common mistakes

  • Incorrectly converting the denominator: A frequent error is writing $\sin^2\theta\cos^2\theta = \sin^2 2\theta$ or $\frac{1}{2}\sin^2 2\theta$, forgetting that the factor of 2 from $\sin 2\theta = 2\sin\theta\cos\theta$ must also be squared, leading to the correct factor of $\frac{1}{4}$.
  • Failure to factorise the numerator: Some candidates correctly get the numerator $\cos^4\theta - \sin^4\theta$ but do not recognise it as a difference of two squares, preventing them from applying the necessary double angle identities.
  • Working from both sides: Students may attempt to manipulate both the LHS and RHS simultaneously to meet in the middle. This is not a valid proof method unless a clear chain of logical equivalence (using $\iff$) is established, which is rare and often done incorrectly.
  • Mixing up identities: Confusing $\cot^2\theta$ with $\cot(2\theta)$ or incorrectly applying the Pythagorean identity, for example by attempting to simplify $\cot^2\theta - \tan^2\theta$ to 1.

  Examiner tip: When proving trigonometric identities, be prepared to use double angle formulae to change the argument of the functions (e.g., from $\theta$ to $2\theta$), especially after combining terms into a single fraction.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.