Solve:tan−1(tan67π)

67π∈(π,23π)So tan−167π=67π−π=6πThe reason we do so is that tan−1m∈(2−π,2π) So we have to find the angle m in (2−π,2π) for which tanm=tan67π

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Solve: { ^{ - 1}}( { {{7 }}{6}} )

Question

Solve:
$tan^{- 1} (tan \frac{7 π}{6})$

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Oliver Discussed

Solve:

tan^{- 1} (tan \frac{7 π}{6})

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9 mins ago

\frac{7 π}{6} \in (π, \frac{3 π}{2})

So tan

^{- 1} \frac{7 π}{6} = \frac{7 π}{6} - π = \frac{π}{6}

The reason we do so is that

t a n^{- 1} m \in (\frac{- π}{2}, \frac{π}{2})

So we have to find the angle m in

(\frac{- π}{2}, \frac{π}{2})

for which

t anm = t an \frac{7 π}{6}

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