Trigonometric Identities

Solution 1: Substituting $\theta = \frac{\pi}{2}$ gives $1 + 0 = 0$ , which is a contradiction. Substituting $\theta = \frac{3\pi}{2}$ gives $-1 + 0 = 0$ , which is a contradiction. Thus $\theta \neq \frac{\pi}{2}, \frac{3\pi}{2}$ , which implies that $\cos \theta \neq 0$ , and so we can divide the equation through by $\cos \theta$ . So, we have $\tan \theta + \sqrt{3} = 0$ $\Rightarrow \tan \theta = -\sqrt{3}$ . Since $\alpha$ and $\beta$ satisfy the given equation, thus $\tan \alpha = \tan \beta = - \sqrt{3}$ .

Using the sum of angles formula for $\tan$ , we have

$\begin{aligned} \tan \left(\alpha + \beta \right) &= \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan\alpha \tan \beta} \\ &= \frac{-\sqrt{3} - \sqrt{3}}{1 - (-\sqrt{3})(-\sqrt{3})} \\ &= \frac{-2\sqrt{3}}{1 - 3} \\ &= \sqrt{3}\\ \end{aligned}$

Hence $\tan^2\left(\alpha + \beta\right)= 3$ .

Solution 2: As shown in Solution 1, we have $\tan \theta = - \sqrt{3}$ .

From the special right triangle $1: \sqrt{3}: 2$ , we have $\tan \left(\frac{\pi}{3}\right) = \sqrt{3}$ . Since $\tan$ is an odd function, so $\tan \left(-\frac{\pi}{3}\right) = -\sqrt{3}$ . $\tan$ also has a period of $\pi$ , so $-\sqrt{3} = \tan \left(-\frac{\pi}{3} + k\pi \right)$ , where $k$ is an integer. Since $0 \leq \theta \leq 2\pi$ , thus $k = 1,2$ are the only possibilities and we get $\alpha = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}$ and $\beta = -\frac{\pi}{3} + 2\pi = \frac{5\pi}{3}$ .

Therefore $\tan\left(\alpha + \beta\right) = \tan\left(\frac{2\pi}{3} + \frac{5\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ . Hence $\tan^2\left(\alpha + \beta\right) = 3$ .