For cos θ ≠ 0, prove that cosec 2θ + cot 2θ = cot θ 9 (b) Explain why cot θ ≠ cosec 2θ + cot 2θ when cos θ = 0 - AQA - A-Level Maths Mechanics - Question 9 - 2020 - Paper 3

To prove the identity, we start by recalling the definitions of the cosecant and cotangent:

• Cosec 2θ is defined as: $\text{cosec } 2\theta = \frac{1}{\sin 2\theta} = \frac{1}{2\sin \theta \cos \theta}$
• Cot 2θ is defined as: $\text{cot } 2\theta = \frac{\cos 2\theta}{\sin 2\theta} = \frac{\cos^{2} \theta - \sin^{2} \theta}{2\sin \theta \cos \theta}$

Substituting these identities into the left hand side gives:

$\text{cosec } 2\theta + \text{cot } 2\theta = \frac{1}{2\sin \theta \cos \theta} + \frac{\cos^{2} \theta - \sin^{2} \theta}{2\sin \theta \cos \theta}$

Combining the terms: $= \frac{1 + \cos^{2} \theta - \sin^{2} \theta}{2\sin \theta \cos \theta}$
Using the identity (1 = \sin^{2} \theta + \cos^{2} \theta), we further simplify to: $= \frac{(\sin^{2} \theta + 2\cos^{2} \theta)}{2\sin \theta \cos \theta}$

Thus, $= \frac{2\cos^{2} \theta}{2\sin \theta \cos \theta} = \frac{\cos \theta}{\sin \theta} = \text{cot } \theta$

This proves the identity.