which of the following is equivalent to $\\tan(\\frac{5\\pi}{6})$? $\\cot(\\frac{5\\pi}{6})$…

Explanation:

Step1: Calculate (\tan(\frac{5\pi}{6}))

We know that (\tan(\alpha)=\frac{\sin\alpha}{\cos\alpha}). For (\alpha = \frac{5\pi}{6}), (\sin(\frac{5\pi}{6})=\frac{1}{2}) and (\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}). So (\tan(\frac{5\pi}{6})=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}})

Step2: Calculate (\cot(\frac{5\pi}{6}))

(\cot(\alpha)=\frac{\cos\alpha}{\sin\alpha}). For (\alpha=\frac{5\pi}{6}), (\cot(\frac{5\pi}{6})=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\neq-\frac{1}{\sqrt{3}})

Step3: Calculate (\tan(-\frac{5\pi}{6}))

Using the property (\tan(-\alpha)=-\tan\alpha), (\tan(-\frac{5\pi}{6})=-\tan(\frac{5\pi}{6})=\frac{1}{\sqrt{3}}\neq-\frac{1}{\sqrt{3}})

Step4: Calculate (\tan(-\frac{\pi}{6}))

Using the property (\tan(-\alpha)=-\tan\alpha), (\tan(-\frac{\pi}{6})=-\tan(\frac{\pi}{6})). And (\tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}), so (\tan(-\frac{\pi}{6})=-\frac{1}{\sqrt{3}})

Step5: Calculate (\tan(\frac{7\pi}{4}))

(\tan(\frac{7\pi}{4})=\tan(2\pi - \frac{\pi}{4})). Using the property (\tan(2\pi-\alpha)=-\tan\alpha), (\tan(\frac{7\pi}{4})=-\tan(\frac{\pi}{4})=- 1\neq-\frac{1}{\sqrt{3}})

Answer:

(\tan(-\frac{\pi}{6}))