The value of tan\(\frac{7\pi}6\) = \(\frac{1}{\sqrt3}\)
∴ The question becomes tan–1\((\frac{1}{\sqrt3})\)
Let,
tan–1\((\frac{1}{\sqrt3})\) = y
⇒ tan y =\((\frac{1}{\sqrt3})\)
⇒ tan\((\frac \pi{6})=(\frac{1}{\sqrt3})\)
The range of the principal value of tan–1 is \((\frac{-\pi}2,\frac{\pi}2)
\) and tan\((\frac \pi{6})=(\frac{1}{\sqrt3})\).
∴ The value of tan–1(tan\(\frac{7\pi}6\)) is \(\frac{\pi}6\).