Using sin² θ + cos² θ = 1, show that the cosec² θ - cot² θ = 1 - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 4

To prove this, we start with the expression:

$ext{cosec}^4 θ - ext{cot}^4 θ$

This can be factored as a difference of squares:

$ext{cosec}^2 θ + ext{cot}^2 θ$

Multiplying by $\left(\text{cosec}^2 θ - ext{cot}^2 θ\right)$ gives us:

$\left(\text{cosec}^2 θ - ext{cot}^2 θ\right)\left(\text{cosec}^2 θ + \text{cot}^2 θ\right) = 1\left(\text{cosec}^2 θ + \text{cot}^2 θ\right)$

Thus:

$ext{cosec}^4 θ - ext{cot}^4 θ = ext{cosec}^2 θ + ext{cot}^2 θ$.

Starting with the equation:

$\text{cosec}^4 θ - \text{cot}^4 θ = 2 - \text{cot} θ,$

we can use the previous result:

$\text{cosec}^4 θ - \text{cot}^4 θ = \text{cosec}^2 θ + \text{cot}^2 θ.$

Thus we have:

$\text{cosec}^2 θ + \text{cot}^2 θ = 2 - \text{cot} θ.$

Rearranging gives:

$\text{cosec}^2 θ + 2\text{cot}^2 θ + \text{cot} θ - 2 = 0.$

Letting $x = \text{cot} θ$, we have:

$2x^2 + x - 1 = 0.$

Using the quadratic formula yields:

$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4}.$

Thus:

1. $\text{cot} θ = \frac{1}{2}$ or
2. $\text{cot} θ = -1.$

Determining angles in the range 90° < θ < 180° gives:

1. $θ = 135°$ for $\text{cot} θ = -1$.