The value of tan−1⁡(tan⁡(−6)),is

The value of $\tan^{- 1} (\tan (- 6))$ ,is

A
$2 π - 6$
B
$2 π + 6$
C
$6 - 2 π$
D
$3 π - 6$

Solution:

We know that $\tan^{- 1} (\tan θ) = θ, if - \frac{π}{2} < θ < \frac{π}{2}$

Here, $θ = - 6$ radians which does not lie between $- \frac{π}{2} and \frac{π}{2}$

However $2 π - 6$ lies between $- \frac{π}{2} and \frac{π}{2}$ such that

$\begin{array}{ll} \tan (2 π - 6) = - \tan 6 = \tan (- 6) \\ ∴ \tan^{- 1} {\tan (- 6)} = \tan^{- 1} {\tan (2 π - 6)} = 2 π - 6 \end{array}$

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