A-Level Mathematics May/June 2025 Q8(b): Hence solve the equation cot²x - tan²x = 5 sec 2x for 0° < x < 90°.

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  The question asks to solve the equation cot²x - tan²x = 5 sec 2x for 0° < x < 90°.

  From part (a), we have the identity cot²x - tan²x ≡ 4 cot 2x cosec 2x.

  Substitute the identity into the equation: $4 \cot 2x \csc 2x = 5 \sec 2x$

  Express all terms using sin 2x and cos 2x: $4 \left( \frac{\cos 2x}{\sin 2x} \right) \left( \frac{1}{\sin 2x} \right) = 5 \left( \frac{1}{\cos 2x} \right)$

  $\frac{4 \cos 2x}{\sin^2 2x} = \frac{5}{\cos 2x}$

  Cross-multiply to eliminate the fractions: $4 \cos^2 2x = 5 \sin^2 2x$

  Divide both sides by cos² 2x to obtain an equation in tan 2x: $4 = 5 \frac{\sin^2 2x}{\cos^2 2x}$

  $4 = 5 \tan^2 2x$

  $\tan^2 2x = \frac{4}{5}$

  Now, solve for tan 2x: $\tan 2x = \pm\sqrt{\frac{4}{5}} = \pm\frac{2}{\sqrt{5}}$

  The given range for x is 0° < x < 90°, so the range for 2x is 0° < 2x < 180°.

  We find the principal value: $\alpha = \tan^{-1}\left(\frac{2}{\sqrt{5}}\right) \approx 41.810°$

  Now we find the solutions for 2x in the range 0° < 2x < 180°.

  Case 1: tan 2x = +\frac{2}{\sqrt{5}} $2x = 41.810°$ (This is in the first quadrant and within our range)

  Case 2: tan 2x = -\frac{2}{\sqrt{5}} $2x = 180° - 41.810° = 138.190°$ (This is in the second quadrant and within our range)

  Now we solve for x: $x = \frac{41.810°}{2} \approx 20.905°$ $x = \frac{138.190°}{2} \approx 69.095°$

  Rounding the solutions to one decimal place: $x = 20.9°$ and $x = 69.1°$

  Both solutions are within the required range 0° < x < 90°.

  Final Answer: $x = 20.9°, 69.1°$

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  How the marks are awarded

  • M1 — For using the identity from part (a) to replace cot²x - tan²x, expressing the equation in terms of sin 2x and cos 2x, and cross-multiplying to obtain an equation like 4 cos² 2x = 5 sin² 2x.
  • A1 — For correctly simplifying the equation to obtain tan² 2x = 4/5 or an equivalent form.
  • DM1 — For the correct method of taking the square root of their tan² 2x = k, finding a principal angle for 2x, and proceeding to find at least one value for x. In this answer, finding x = 20.9° earns this mark.
  • A1 — For obtaining both correct solutions, x = 20.9° and x = 69.1°, and no others within the given range.

  Common mistakes

  • Forgetting to consider the negative square root when solving tan² 2x = 4/5, which results in missing the second solution.
  • Incorrectly managing the interval. Forgetting to change the range from 0° < x < 90° to 0° < 2x < 180° can lead to finding incorrect solutions or missing correct ones.
  • Attempting to solve the equation without using the identity from part (a), which is much more complicated and does not follow the instruction 'Hence'.
  • Making an algebraic error after substitution, such as incorrectly simplifying (cos 2x / sin 2x) * (1 / sin 2x).

  Examiner tip: The word 'Hence' is a direct instruction to use the result from the previous part of the question; building this connection is the key to unlocking the simplest solution path.

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