Question

which of the following is equivalent to \\(\tan\left(\frac{5\pi}{6}\
ight)\\)? \\(\tan\left(\frac{7\pi}{6}\
ight)\\) \\(\tan\left(-\frac{5\pi}{6}\
ight)\\) \\(\tan\left(-\frac{\pi}{6}\
ight)\\) \\(\cot\left(\frac{5\pi}{6}\
ight)\\)

Explanation:

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

We know that \(\tan(A + B)=\frac{\tan A+\tan B}{1 - \tan A\tan B}\), and \(\frac{5\pi}{6}=\pi-\frac{\pi}{6}\).
Using the formula \(\tan(\pi - \alpha)=-\tan\alpha\), so \(\tan(\frac{5\pi}{6})=\tan(\pi-\frac{\pi}{6})=-\tan\frac{\pi}{6}=-\frac{\sqrt{3}}{3}\).

Step2: Calculate \(\tan(\frac{7\pi}{6})\)

\(\frac{7\pi}{6}=\pi+\frac{\pi}{6}\), using the formula \(\tan(\pi+\alpha)=\tan\alpha\), so \(\tan(\frac{7\pi}{6})=\tan(\pi + \frac{\pi}{6})=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}\).

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

Using the formula \(\tan(-\alpha)=-\tan\alpha\), \(\tan(-\frac{5\pi}{6})=-\tan\frac{5\pi}{6}=-(-\frac{\sqrt{3}}{3})=\frac{\sqrt{3}}{3}\).

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

Using the formula \(\tan(-\alpha)=-\tan\alpha\), \(\tan(-\frac{\pi}{6})=-\tan\frac{\pi}{6}=-\frac{\sqrt{3}}{3}\).

Step5: Calculate \(\cot(\frac{5\pi}{6})\)

We know that \(\cot\alpha=\frac{1}{\tan\alpha}\), \(\cot(\frac{5\pi}{6})=\frac{1}{\tan(\frac{5\pi}{6})}=\frac{1}{-\frac{\sqrt{3}}{3}}=-\sqrt{3}\).

Answer:

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