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(a) Prove that $$ tan\theta + cot\theta = 2cosec2\theta, \quad \theta \neq \frac{n\pi}{2}, \quad n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

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(a)-Prove-that--$$-tan\theta-+-cot\theta-=-2cosec2\theta,-\quad-\theta-\neq-\frac{n\pi}{2},-\quad-n-\in-\mathbb{Z}-Edexcel-A-Level Maths Pure-Question 10-2017-Paper 1.png

(a) Prove that $$ tan\theta + cot\theta = 2cosec2\theta, \quad \theta \neq \frac{n\pi}{2}, \quad n \in \mathbb{Z}. $$ (b) Hence explain why the equation $$ tan\th... show full transcript

Worked Solution & Example Answer:(a) Prove that $$ tan\theta + cot\theta = 2cosec2\theta, \quad \theta \neq \frac{n\pi}{2}, \quad n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Step 1

Prove that $tan\theta + cot\theta = 2cosec2\theta$

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Answer

To prove the identity, we start by rewriting the left-hand side:

  1. Write the definitions of tangent and cotangent: tanθ=sinθcosθ,cotθ=cosθsinθtan\theta = \frac{sin\theta}{cos\theta}, \quad cot\theta = \frac{cos\theta}{sin\theta}
  2. Combine the two terms: tanθ+cotθ=sinθcosθ+cosθsinθtan\theta + cot\theta = \frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} =sin2θ+cos2θsinθcosθ= \frac{sin^2\theta + cos^2\theta}{sin\theta cos\theta}
  3. Recognize that sin2θ+cos2θ=1sin^2\theta + cos^2\theta = 1: =1sinθcosθ= \frac{1}{sin\theta cos\theta}
  4. Using the double angle identity, we know: sin2θ=2sinθcosθsin 2\theta = 2sin\theta cos\theta So, 1sinθcosθ=2sin2θ\frac{1}{sin\theta cos\theta} = \frac{2}{sin 2\theta}
  5. Substituting this into the equation: tanθ+cotθ=2cosec2θtan\theta + cot\theta = 2cosec2\theta This proves part (a) of the question.

Step 2

Explain why $tan\theta + cot\theta = 1$ does not have any real solutions

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Answer

From the derived equation, we see that: tanθ+cotθ=2cosec2θtan\theta + cot\theta = 2cosec2\theta If we set this equal to 1: 2cosec2θ=1cosec2θ=122cosec2\theta = 1 \Rightarrow cosec2\theta = \frac{1}{2}

This implies: sin2θ=2sin2\theta = 2 However, the sine function has a range of [-1, 1]. Therefore, the statement 22 outside this range shows that there are no possible angles heta heta that satisfy the equation. Hence, there are no real solutions for tanθ+cotθ=1tan\theta + cot\theta = 1.

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