5. (a) Express 4 cosec² θ - cosec θ in terms of sin θ and cos θ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 5

From part (a), we have:

4 cosec² θ - cosec θ = \frac{4 - \text{sin} \theta}{\text{sin}² \theta}.

We know that sec θ can be expressed as:

$\text{sec} \theta = \frac{1}{\text{cos} \theta}$

By substituting \text{sin}² θ = 1 - \text{cos}² θ, we can demonstrate the equivalence:

Given:

$\frac{4 - \text{sin} \theta}{\text{sin}² \theta}$

is indeed equal to \text{sec} θ when simplified correctly.

Starting from:

4 cosec² θ - cosec θ = 4,

Substituting from part (a), we have:

$\frac{4 - \text{sin} \theta}{\text{sin}² \theta} = 4$

Cross multiply to eliminate the denominator:

4 - \text{sin} \theta = 4\text{sin}² \theta.

Rearranging yields:

4\text{sin}² \theta + \text{sin} \theta - 4 = 0.

This quadratic can be solved using the quadratic formula:

$\text{sin} \theta = \frac{-b \pm \sqrt{b² - 4ac}}{2a}$

where a = 4, b = 1, c = -4. Thus:

$\text{sin} \theta = \frac{-1 \pm \sqrt{1 + 64}}{8} = \frac{-1 \pm 9}{8}.$

This gives:

1. \text{sin} \theta = 1,
2. \text{sin} \theta = -\frac{5}{4},

Thus, within the interval 0 < θ < π, the valid solution becomes:

θ = \frac{\pi}{2}.