find the exact value of \\( \\tan ^ { - 1 } \\left( \\tan \\left( - \\frac { 7 \\pi } { 5 } \\right)…

find the exact value of \\( \\tan ^ { - 1 } \\left( \\tan \\left( - \\frac { 7 \\pi } { 5 } \\right) \\right) \\).\nwrite your answer in radians in terms of \\( \\pi \\).\nif necessary, click on \undefined.\

find the exact value of \\( \\tan ^ { - 1 } \\left( \\tan \\left( - \\frac { 7 \\pi } { 5 } \\right) \\right) \\).\nwrite your answer in radians in terms of \\( \\pi \\).\nif necessary, click on \undefined.\

Answer

Explanation:

Step1: Use the periodicity of the tangent function

The tangent function (y = \tan(x)) has a period of (\pi), so (\tan\left(-\frac{7\pi}{5}\right)=\tan\left(-\frac{7\pi}{5}+ 2\pi\right)). [ -\frac{7\pi}{5}+2\pi=-\frac{7\pi}{5}+\frac{10\pi}{5}=\frac{3\pi}{5} ]

Step2: Consider the range of the inverse - tangent function

The range of (y = \tan^{-1}(x)) is (\left(-\frac{\pi}{2},\frac{\pi}{2}\right)). Since (\tan\left(\frac{3\pi}{5}\right)=\tan\left(\pi - \frac{2\pi}{5}\right)=-\tan\left(\frac{2\pi}{5}\right)) and (\tan^{-1}(\tan(x))=x) when (x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)) (\tan^{-1}\left(\tan\left(-\frac{7\pi}{5}\right)\right)=\tan^{-1}\left(\tan\left(-\frac{7\pi}{5}+2\pi\right)\right)=\tan^{-1}\left(\tan\left(\frac{3\pi}{5}\right)\right)) (\tan\left(\frac{3\pi}{5}\right)=\tan\left(-\frac{2\pi}{5}+\pi\right)), and (\tan^{-1}(\tan(x))) for (x =-\frac{2\pi}{5}) (because (-\frac{2\pi}{5}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)))

Answer:

(-\frac{2\pi}{5})