A-level · Trigonometry

The sec/cosec/cot identitiesExam-style questions

The two new Pythagorean identities 1 + tan²θ ≡ sec²θ and 1 + cot²θ ≡ cosec²θ — deriving them from sin²+cos²≡1, and using them to prove identities, simplify expressions and turn equations into a solvable quadratic.

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17 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAllLevel 2Level 3Level 4Level 5
  1. 1
    Solve, for 0<x<360∘0 < x < 360^\circ, the equation

    2cot⁡2x+5cosec⁡x+1=02\cot^2 x + 5\operatorname{cosec} x + 1 = 0,

    giving your answers to 1 decimal place. You must show that one possible value of cosec⁡x\operatorname{cosec} x gives no solutions, with a reason. (7 marks)
    Level 4 · 7 marksTry it
  2. 2
    (a) Show that the expression

    5sin⁡2θsec⁡θ+3cos⁡2θ cosec⁡ θ5\sin 2\theta \sec\theta + 3\cos 2\theta \, \operatorname{cosec}\,\theta

    can be written as 4sin⁡θ+3cosec⁡ θ4\sin\theta + 3\operatorname{cosec}\,\theta for values of θ\theta at which sin⁡θ\sin\theta and cos⁡θ\cos\theta are both non-zero.

    (4 marks)
    (b) The equation

    5sin⁡2θsec⁡θ+3cos⁡2θ cosec⁡ θ=−75\sin 2\theta \sec\theta + 3\cos 2\theta \, \operatorname{cosec}\,\theta = -7

    is to be solved for 0°≤θ≤360°0° \le \theta \le 360°

    A student's solution, which uses the result of part (a), is shown below.

    Line 1 4sin⁡θ+3cosec⁡ θ=−7\quad 4\sin\theta + 3\operatorname{cosec}\,\theta = -7

    Line 2 4sin⁡θ+3sin⁡θ+7=0\quad 4\sin\theta + \dfrac{3}{\sin\theta} + 7 = 0

    Line 3 4sin⁡2θ+7sin⁡θ+3=0\quad 4\sin^2\theta + 7\sin\theta + 3 = 0

    Line 4 sin⁡θ=−1\quad \sin\theta = -1 or sin⁡θ=−0.75\sin\theta = -0.75

    Line 5 θ=228.6°, 270°, 311.4°\quad \theta = 228.6°,\ 270°,\ 311.4°
    (b) (i) The student should not have kept both values of sin⁡θ\sin\theta in Line 4.

    Explain why one of them must be rejected.

    (ii) State all the solutions of the original equation in the interval 0°≤θ≤360°0° \le \theta \le 360°

    (2 marks)
    Level 4 · 6 marksTry it
  3. 3
    In this question you must show all stages of your working.

    Solutions relying entirely on calculator technology are not acceptable.
    (a) Prove that

    cot⁡θcosec⁡θ−1−cot⁡θcosec⁡θ+1≡2tan⁡θθ≠nπ2, n∈Z\frac{\cot\theta}{\operatorname{cosec}\theta - 1} - \frac{\cot\theta}{\operatorname{cosec}\theta + 1} \equiv 2\tan\theta \qquad \theta \neq \frac{n\pi}{2},\ n \in \mathbb{Z}

    (3 marks)
    (b) Hence solve, for 0<x<180∘0 < x < 180^\circ, the equation

    4cot⁡xcosec⁡x−1−4cot⁡xcosec⁡x+1=4sec⁡2x−1\frac{4\cot x}{\operatorname{cosec} x - 1} - \frac{4\cot x}{\operatorname{cosec} x + 1} = 4\sec^2 x - 1

    Give each answer, in degrees, to one decimal place.

    (4 marks)
    Level 4 · 7 marksTry it
  4. 4
    (a) Prove that 1+sec⁡θtan⁡θ+sin⁡θ≡cosec⁡θ\dfrac{1 + \sec\theta}{\tan\theta + \sin\theta} \equiv \operatorname{cosec}\theta.

    (4 marks)
    (b) In this question you must show detailed reasoning.

    Hence solve, for 0<θ<2π0 < \theta < 2\pi, the equation 1+sec⁡θtan⁡θ+sin⁡θ=6−4sin⁡θ.\frac{1 + \sec\theta}{\tan\theta + \sin\theta} = 6 - 4\sin\theta. Give your answers correct to 3 decimal places.

    (5 marks)
    Level 4 · 9 marksTry it
  5. 5
    In this question you must show detailed reasoning.
    (a) Show that the equation 9cosec⁡θ−8sin⁡θ=kcos⁡θ9\operatorname{cosec}\theta - 8\sin\theta = k\cos\theta, where kk is a constant, can be written as 9cot⁡2θ−kcot⁡θ+1=0.9\cot^2\theta - k\cot\theta + 1 = 0.

    (3 marks)
    (b) For a particular positive value of kk, the equation 9cosec⁡θ−8sin⁡θ=kcos⁡θ9\operatorname{cosec}\theta - 8\sin\theta = k\cos\theta has exactly one root in the interval 0∘<θ<180∘0^\circ < \theta < 180^\circ.

    Find this root, correct to 3 significant figures.

    (5 marks)
    Level 4 · 8 marksTry it