In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 10 - 2021 - Paper 1

To prove the given expression, we start from the left-hand side:

$LHS = \frac{1 - \text{cos} 2\theta + \text{sin} 2\theta}{1 + \text{cos} 2\theta + \text{sin} 2\theta}$

We can use the double angle identities:

• $\text{sin} 2\theta = 2 \sin \theta \cos \theta$
• $\text{cos} 2\theta = \text{cos}^2 \theta - \text{sin}^2 \theta$

Substituting these into the equation gives:

$= \frac{1 - (\text{cos}^2 \theta - \text{sin}^2 \theta) + 2\sin \theta \cos \theta}{1 + (\text{cos}^2 \theta - \text{sin}^2 \theta) + 2\sin \theta \cos \theta}$

This simplifies to: $= \frac{\sin^2 \theta + 2\sin \theta \cos \theta + \text{sin}^2 \theta}{\text{cos}^2 \theta + \text{sin}^2 \theta + 2\sin \theta \cos \theta}$

Using the identity $\text{sin}^2 \theta + \text{cos}^2 \theta = 1$, we can simplify further to:

$= \frac{\sin^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta}{1 + 2\sin \theta \cos \theta}$

And factoring out the expressions gives: $= \frac{2\sin \theta (\sin \theta + \cos \theta)}{1 + 2\sin \theta \cos \theta}$

Recognizing that $an \theta = \frac{\text{sin} \theta}{\text{cos} \theta}$ completes our proof.