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Trigonometric Functions

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Question 104
a.

Show that sec⁡θsec⁡θ−cos⁡θ=11−cos⁡2θ\displaystyle \frac{\sec \theta}{\sec \theta - \cos \theta} = \frac{1}{1 - \cos^2 \theta}secθ−cosθsecθ​=1−cos2θ1​

[2]
b.

Hence prove that sec⁡θsec⁡θ−cos⁡θ≡csc⁡2θ\displaystyle \frac{\sec \theta}{\sec \theta - \cos \theta} \equiv \csc^2 \thetasecθ−cosθsecθ​≡csc2θ

[2]
c.

Hence solve, for 0<θ<2π0 < \theta < 2\pi0<θ<2π, sec⁡θsec⁡θ−cos⁡θ=2cot⁡θ\displaystyle \frac{\sec \theta}{\sec \theta - \cos \theta} = 2 \cot \thetasecθ−cosθsecθ​=2cotθ

[3]
Markscheme

Trigonometric Functions Questions

  1. A Level
  2. /Maths
  3. /Trigonometric Functions

274 exam-style questions on Edexcel A Level Maths Trigonometric Functions, covering 6.1 Secant, Cosecant and Cotangent, 6.2 Graphs of Sec x, Cosec x and Cot x, 6.3 Using Sec x, Cosec x and Cot x, 6.4 Trigonometric Identities, and 6.5 Inverse Trigonometric Functions. Each one has a worked solution and a mark scheme showing where the marks go.

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