Using \( \sin^2 \theta + \cos^2 \theta = 1 \), show that the \( \csc^2 \theta - \cot^2 \theta = 1 \) - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 4

From the previous step, we already have:
$\csc^2 \theta = 1 + \cot^2 \theta$
Now, substituting this into ( \csc^2 \theta - \cot^4 \theta ):
$\csc^2 \theta - \cot^4 \theta = (1 + \cot^2 \theta) - \cot^4 \theta$
This can be rearranged to:
$1 + \cot^2 \theta - \cot^4 \theta \quad (1)$
Noticing that ( \cot^4 \theta = (\cot^2 \theta)^2 ) allows us to use factoring techniques to show that:
$\csc^2 \theta + \cot^2 \theta = 1 + 2 \cot^2 \theta - \cot^4 \theta$
Thus proving the required result.

Starting with the equation:
$\csc^4 \theta - \cot^4 \theta = 2 - \cot \theta$
Using the identity obtained earlier, we can reformulate this into a quadratic in terms of ( \cot \theta ):
Let ( x = \cot \theta ):
$\csc^4 \theta = (1+x^2)^2 - x^4$
Thus, we have the quadratic equation:
$1 + 2x^2 - x^4 - 2 + x = 0$
Which simplifies to:
$-x^4 + 2x^2 + x - 1 = 0$
Solving for x gives possible values for ( \theta ). Lastly, we confirm that ( \theta = 135^{\circ} ) satisfies this equation, considering the constraints.