Mathematics 9709/22 — May/June 2023

Walkthrough

The equation contains both $\sec\theta$ and $\tan\theta$. To solve it, we want a single trigonometric function. The identity $\tan^2\theta = \sec^2\theta - 1$ is exactly the link we need, because it lets us replace every $\tan^2\theta$ with an expression in $\sec^2\theta$. After substituting and simplifying, the equation becomes a quadratic in $\sec\theta$.

Next, solve the quadratic. Factorising gives two possible values for $\sec\theta$: $-\frac{2}{3}$ and $\frac{7}{2}$. Since $\sec\theta = \frac{1}{\cos\theta}$, these correspond to $\cos\theta = -\frac{3}{2}$ and $\cos\theta = \frac{2}{7}$. The first is impossible because the cosine of any real angle must lie between $-1$ and $1$. So only $\cos\theta = \frac{2}{7}$ remains.

Finally, find all angles in $0^\circ < \theta < 360^\circ$ with $\cos\theta = \frac{2}{7}$. Cosine is positive in the first and fourth quadrants, so the two solutions are the principal value $\cos^{-1}(\frac{2}{7})$ and its reflection $360^\circ - \cos^{-1}(\frac{2}{7})$. These give approximately $73.4^\circ$ and $286.6^\circ$.

Key Takeaways

• The Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ reduces equations involving both $\sec$ and $\tan$ to a single variable.
• A quadratic in $\sec\theta$ can be solved like any quadratic, then converted to $\cos\theta$ using $\sec\theta = \frac{1}{\cos\theta}$.
• Always check that possible values of $\cos\theta$ lie in $[-1, 1]$; impossible roots must be discarded.
• For $\cos\theta = k$ with $0 < k < 1$, the solutions in $0^\circ$ to $360^\circ$ are $\cos^{-1}k$ and $360^\circ - \cos^{-1}k$.

Common Mistakes

• Forgetting to multiply the $5$ through the bracket: $5(\sec^2\theta - 1) = 5\sec^2\theta - 5$, not $5\sec^2\theta - 1$.
• Making a sign error when rearranging to form the quadratic.
• Factorising $6x^2 - 17x - 14$ incorrectly; check by expanding $(3x+2)(2x-7)$.
• Forgetting that $\sec\theta = \frac{1}{\cos\theta}$, so $\sec\theta = -\frac{2}{3}$ means $\cos\theta = -\frac{3}{2}$, which is invalid.
• Giving only the acute solution $73.4^\circ$ and missing the fourth-quadrant solution $286.6^\circ$.
• Working in radians when the question asks for degrees, or giving angles outside the interval $0^\circ < \theta < 360^\circ$.

Things to Be Careful About

The domain is $0^\circ < \theta < 360^\circ$, so do not include $0^\circ$ or $360^\circ$ themselves. The final answers should be given in degrees. The mark scheme accepts $73.4^\circ$ and $286.6^\circ$ or greater accuracy, and requires no other solutions in the interval. Since $\cos\theta = \frac{2}{7}$ has exactly two solutions in the interval, and the other root is impossible, these are the only answers.