8. (a) Prove that sec 2A + tan 2A = \frac{cos A + sin A}{cos A - sin A} ; \quad A \neq \frac{(2n + 1)\pi}{4}, \ n \in \mathbb{Z} (b) Hence solve, for 0 \leq \theta < 2\pi, sec 2\theta + tan 2\theta = \frac{1}{2} Give your answers to 3 decimal places. - Edexcel - A-Level Maths Pure - Question 9 - 2015 - Paper 3

To prove the identity, we start with the left side:

$sec \, 2A + tan \, 2A = \frac{1}{cos \, 2A} + \frac{sin \, 2A}{cos \, 2A} = \frac{1 + sin \, 2A}{cos \, 2A}$

Next, we use the double angle identities:

$sin \, 2A = 2 sin \, A cos \, A$
$cos \, 2A = cos^2 \, A - sin^2 \, A$

Substituting these into the equation gives us:

$\frac{1 + 2 sin \, A cos \, A}{cos^2 \, A - sin^2 \, A}$

We can rewrite the denominator using the identity:

$cos^2 \, A - sin^2 \, A = (cos \, A + sin \, A)(cos \, A - sin \, A)$

This allows us to express the left side as:

$\frac{(1 + 2 sin \, A cos \, A)}{(cos \, A + sin \, A)(cos \, A - sin \, A)}$

Simplifying this leads us to show that it is equal to the right side. Canceling out common terms results in the required identity effectively proving that:

$sec \, 2A + tan \, 2A = \frac{cos \, A + sin \, A}{cos \, A - sin \, A}$.