5. (a) Express $4\csc^2 \theta - \csc^2 \theta$ in terms of $\sin \theta$ and $\cos \theta$ - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 5

From part (a), we found:

$4\csc^2 \theta - \csc^2 \theta = \frac{3}{\sin^2 \theta}$

Now, we need to show that:

$\frac{3}{\sin^2 \theta} = \sec \theta$

Using the definition of secant:

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

And since $\sin^2 \theta + \cos^2 \theta = 1$, we have:

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

Now, if we manipulate this, we can express $\sec \theta$ in relation to $\sin \theta$ as follows:

$\sec \theta = \frac{3}{\sin^2 \theta}$

First, we have established that:

$4\csc^2 \theta - \csc^2 \theta = \frac{3}{\sin^2 \theta}$

Setting this equal to 4 gives:

$\frac{3}{\sin^2 \theta} = 4$

Cross multiplying leads to:

\sin^2 \theta = \frac{3}{4} \sin \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ This implies: $$\theta = \frac{\pi}{3} , \frac{2\pi}{3}$$ These solutions fall within the specified range of $0 < \theta < \pi$.