(a) Prove that
$$
tan\theta + cot\theta = 2cosec2\theta, \quad \theta \neq \frac{n\pi}{2}, \quad n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1
Question 10
(a) Prove that
$$
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:
Write the definitions of tangent and cotangent:
tanθ=cosθsinθ,cotθ=sinθcosθ
Combine the two terms:
tanθ+cotθ=cosθsinθ+sinθcosθ=sinθcosθsin2θ+cos2θ
Recognize that sin2θ+cos2θ=1:
=sinθcosθ1
Using the double angle identity, we know:
sin2θ=2sinθcosθ
So,
sinθcosθ1=sin2θ2
Substituting this into the equation:
tanθ+cotθ=2cosec2θ
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=2cosec2θ
If we set this equal to 1:
2cosec2θ=1⇒cosec2θ=21
This implies:
sin2θ=2
However, the sine function has a range of [-1, 1]. Therefore, the statement 2 outside this range shows that there are no possible angles heta that satisfy the equation. Hence, there are no real solutions for tanθ+cotθ=1.