8. (a) Starting from the formulae for sin(A + B) and cos(A + B), prove that $$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$ (b) Deduce that $$\tan\left(\theta + \frac{\pi}{6}\right) = \frac{1 + \sqrt{3} \tan \theta}{\sqrt{3} - \tan \theta}$$ (c) Hence, or otherwise, solve, for $0 \leq \theta < \pi$, $$1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta) \tan(\pi - \theta)$$ Give your answers as multiples of $\pi$ - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 6

To deduce this identity, we apply the expression for $\tan(A + B)$ using angles relevant to our problem:

Let $A = \theta$ and $B = \frac{\pi}{6}$. First, compute $\tan \frac{\pi}{6}$, which is known to be:

$\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$

Now substitute this into the equation:

$\tan\left(\theta + \frac{\pi}{6}\right) = \frac{\tan \theta + \tan \frac{\pi}{6}}{1 - \tan \theta \tan \frac{\pi}{6}}$

This simplifies to:

$\tan\left(\theta + \frac{\pi}{6}\right) = \frac{\tan \theta + \frac{1}{\sqrt{3}}}{1 - \tan \theta \cdot \frac{1}{\sqrt{3}}}$

Multiplying both the numerator and denominator by $\sqrt{3}$, we obtain:

$\tan\left(\theta + \frac{\pi}{6}\right) = \frac{\sqrt{3} \tan \theta + 1}{\sqrt{3} - \tan \theta}$

Thus, the identity is deduced.

Recognizing that $\tan(\pi - \theta) = -\tan \theta$, we can rewrite the equation as:

$1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta)(-\tan \theta)$

This expands to:

$1 + \sqrt{3} \tan \theta = -\sqrt{3} \tan \theta + \tan^2 \theta$

Rearranging gives:

$\tan^2 \theta + (\sqrt{3} + \sqrt{3}) \tan \theta - 1 = 0$

This is a standard quadratic equation in terms of $\tan \theta$:

$\tan^2 \theta + 2\sqrt{3} \tan \theta - 1 = 0$

Using the quadratic formula:

$\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 2\sqrt{3}$, and $c = -1$. Then we calculate:

$\tan \theta = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4(1)(-1)}}{2(1)}$ $\tan \theta = \frac{-2\sqrt{3} \pm \sqrt{12 + 4}}{2}$ $\tan \theta = \frac{-2\sqrt{3} \, \pm \sqrt{16}}{2}$ $\tan \theta = \frac{-2\sqrt{3} \, \pm 4}{2}$

Thus, we find:

1. $\tan \theta = 2 - \sqrt{3}$
2. $\tan \theta = -\sqrt{3} - 2$

Considering the domain $0 \leq \theta < \pi$, we compute:

• For $\tan \theta = 2 - \sqrt{3}$, we find $\theta = \tan^{-1}(2 - \sqrt{3})$.
• The second solution falls outside the valid domain since it gives negative values for $\tan \theta$.

Finally, the solution is given in terms of multiples of $\pi$ as $\theta = \tan^{-1}(2 - \sqrt{3}) + n\pi$, where $n = 0$.