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Secant, cosecant and cotangentEdexcel International A Level Maths: Flashcards

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Question

Define $\sec x$.

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Define sec⁡x\sec x.
sec⁡x=1cos⁡x\sec x=\frac{1}{\cos x}, undefined where cos⁡x=0\cos x=0.
Define cosec⁡ x\operatorname{cosec}\,x.
cosec⁡ x=1sin⁡x\operatorname{cosec}\,x=\frac{1}{\sin x}, undefined where sin⁡x=0\sin x=0.
Define cot⁡x\cot x in two ways.
cot⁡x=1tan⁡x=cos⁡xsin⁡x\cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}.
Where are the asymptotes of y=sec⁡xy=\sec x?
At x=π2+nπx=\frac{\pi}{2}+n\pi (where cos⁡x=0\cos x=0).
Where are the asymptotes of y=cosec⁡ xy=\operatorname{cosec}\,x and y=cot⁡xy=\cot x?
At x=nπx=n\pi (where sin⁡x=0\sin x=0).
Range of sec⁡x\sec x and cosec⁡ x\operatorname{cosec}\,x?
y≤−1y\le-1 or y≥1y\ge1; no values between −1-1 and 11.
Range of cot⁡x\cot x?
All real numbers.
Period of sec⁡x\sec x, cosec⁡ x\operatorname{cosec}\,x, cot⁡x\cot x?
2π2\pi, 2π2\pi and π\pi respectively.
Restricted domain used for sec⁡x\sec x?
0≤x≤π0\le x\le\pi, x≠π2x\ne\frac{\pi}{2}.
Restricted domain used for cosec⁡ x\operatorname{cosec}\,x?
−π2≤x≤π2-\frac{\pi}{2}\le x\le\frac{\pi}{2}, x≠0x\ne0.
Restricted domain used for cot⁡x\cot x?
0<x<π0<x<\pi.
Exact values of sec⁡60∘\sec60^{\circ} and cot⁡45∘\cot45^{\circ}?
22 and 11.
Does sec⁡θ=12\sec\theta=\frac12 have any solutions?
No, since ∣sec⁡θ∣≥1|\sec\theta|\ge1.

Exam questions on Secant, cosecant and cotangent

  1. The angle θ\theta is acute and sin⁡θ=513\sin\theta=\frac{5}{13}.
    Find the exact value of sec⁡θ+cosec⁡ θ\sec\theta+\operatorname{cosec}\,\theta.2 marks
  2. Throughout, xx is measured in radians, and each function is considered only where it is defined.
    Write cosec⁡ xcot⁡x\dfrac{\operatorname{cosec}\,x}{\cot x} as a single trigonometric function.2 marks
  3. Angles are measured in degrees and 0≤θ<3600\le\theta<360.
    Solve sec⁡θ=−2\sec\theta=-2.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).