6. (a) Prove that $$\frac{1}{\sin 2\theta} = \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, \quad \theta \neq 90^\circ, n \in \mathbb{Z}$$ (b) Hence, or otherwise, (i) show that $\tan 15^\circ = 2 - \sqrt{3},$ (ii) solve, for $0 < x < 360^\circ,$ $$\csc 4x - \cot 4x = 1$$ - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 3

To show this result, we can utilize the tangent subtraction formula:

1. Using the formula for tangent of a difference: $\tan(60^\circ - 45^\circ) = \frac{\tan 60^\circ - \tan 45^\circ}{1 + \tan 60^\circ \tan 45^\circ}$

2. Substituting known values: $\tan 60^\circ = \sqrt{3},\;\tan 45^\circ = 1$ $\tan 15^\circ = \frac{\sqrt{3} - 1}{1 + \sqrt{3}}$

3. Rationalizing the denominator: $= \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{2 - \sqrt{3}}{2 - 3} = 2 - \sqrt{3}$

Thus, confirming that (\tan 15^\circ = 2 - \sqrt{3}).

To solve the equation:

1. Rearranging the equation, we have: $\csc 4x = \cot 4x + 1$

2. Replacing cosecant and cotangent with sine and cosine: $\frac{1}{\sin 4x} = \frac{\cos 4x}{\sin 4x} + 1$

3. This simplifies to: $\frac{1}{\sin 4x} = \frac{\cos 4x + \sin 4x}{\sin 4x}$

4. Thus, multiplying through by (\sin 4x) leads to: $1 = \cos 4x + \sin 4x$

5. Rearranging gives: $\cos 4x + \sin 4x = 1$

It can be solved further using trigonometric identities such as the unit circle or additional transformations, providing the specific values of (x) within the range ((0, 360)).