Using sin² θ + cos² θ = 1, show that the cosec² θ - cot² θ = 1 - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 4

Using the identity derived from part (a), we can express cosec⁴ θ - cot⁴ θ in terms of cosec² θ and cot² θ. We utilize the difference of squares formula:

$cosec⁴ θ - cot⁴ θ = (cosec² θ - cot² θ)(cosec² θ + cot² θ)$

From part (a), we know that:

$cosec² θ - cot² θ = 1$

Therefore, substituting in gives us:

$cosec⁴ θ - cot⁴ θ = 1(cosec² θ + cot² θ) = cosec² θ + cot² θ$

Starting with the equation from part (c):

$cosec⁴ θ - cot⁴ θ = 2 - cot θ$

Substituting the expression from part (b) gives:

$cosec² θ + cot² θ = 2 - cot θ$

Now rearranging for terms involving cot θ leads to:

$cot² θ + cot θ + 1 - 2 = 0$

Which simplifies to:

$cot² θ + cot θ - 1 = 0$

We can solve this quadratic equation using the quadratic formula:

$cot θ = \frac{-b \pm \sqrt{b² - 4ac}}{2a}$

Here, a = 1, b = 1, and c = -1:

$cot θ = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$

Calculating gives us two possible solutions for cot θ:

$cot θ = \frac{-1 + \sqrt{5}}{2} \text{ or } cot θ = \frac{-1 - \sqrt{5}}{2}$

Considering the range 90° < θ < 180°, we find:

$θ = 135°$