what is the exact value of tan(300°)?\n-√3\n-...

# Explanation: ## Step1: Rewrite the angle We know that \(300^{\circ}=360^{\circ}- 60^{\circ}\), so \(\tan(300^{\circ})=\tan(360^{\circ}-60^{\circ})\). ## Step2: Use the tangent - angle formula The formula for \(\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}\), and for \(\tan(360^{\circ}-60^{\circ})\), since \(\tan(360^{\circ}) = 0\) and \(\tan(60^{\circ})=\sqrt{3}\), we have \(\tan(360^{\circ}-60^{\circ})=\frac{\tan360^{\circ}-\tan60^{\circ}}{1+\tan360^{\circ}\tan60^{\circ}}=\frac{0 - \sqrt{3}}{1+0\times\sqrt{3}}=-\sqrt{3}\). Another way is to use the unit - circle definition. The angle \(300^{\circ}\) is in the fourth quadrant, and the reference angle is \(60^{\circ}\). In the fourth quadrant, the tangent function is negative, and \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), for \(\theta = 300^{\circ}\), \(\sin(300^{\circ})=-\frac{\sqrt{3}}{2}\) and \(\cos(300^{\circ})=\frac{1}{2}\), so \(\tan(300^{\circ})=\frac{\sin(300^{\circ})}{\cos(300^{\circ})}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\). # Answer: A. \(-\sqrt{3}\)