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Question: Find the value of each of the following: 1)\({\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\...

Find the value of each of the following:
1)tan⁡−1(13){\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)
2)tan⁡−1(−13){\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{\sqrt 3 }}} \right)
3)tan⁡−1(cos⁡(π2)){\tan ^{ - 1}}\left( {\cos \left( {\dfrac{\pi }{2}} \right)} \right)
4)tan⁡−1(2cos⁡(2π3)){\tan ^{ - 1}}\left( {2\cos \left( {\dfrac{{2\pi }}{3}} \right)} \right)

Explanation

Solution

To find the solution of each of these inverse tangent functions, we have to find the value of the tangent function of the values in the bracket, because we know,
tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta
And, for those, in which the cosine functions are given separately inside the bracket, we have to change the cosine function into tangent function using the triangle formula and use required operations to find the required values.

Complete step-by-step solution:

  1. tan⁡−1(13){\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)
    Now, we know, tan⁡π6=13\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}
    Replacing this in the given function, we get,
    tan⁡−1(tan⁡π6){\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{6}} \right)
    Now, we know, tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta .
    Therefore, using this trigonometric property, we get,
    tan⁡−1(tan⁡π6)=π6{\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{6}} \right) = \dfrac{\pi }{6}
  2. tan⁡−1(−13){\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{\sqrt 3 }}} \right)
    We know, the principle domain of tan⁡−1{\tan ^{ - 1}} lies in (−π2,π2)\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right).
    So, we know, tan⁡(−π6)=−13\tan \left( {\dfrac{{ - \pi }}{6}} \right) = \dfrac{{ - 1}}{{\sqrt 3 }}.
    Therefore, we can write the inverse function as,
    tan⁡−1(tan⁡(−π6)){\tan ^{ - 1}}\left( {\tan \left( {\dfrac{{ - \pi }}{6}} \right)} \right)
    Now, using the property tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta , we get,
    tan⁡−1(tan⁡(−π6))=−π6{\tan ^{ - 1}}\left( {\tan \left( {\dfrac{{ - \pi }}{6}} \right)} \right) = \dfrac{{ - \pi }}{6}
  3. tan⁡−1(cos⁡(π2)){\tan ^{ - 1}}\left( {\cos \left( {\dfrac{\pi }{2}} \right)} \right)
    We know, cos⁡π2=0\cos \dfrac{\pi }{2} = 0.
    So, tan⁡−1(0){\tan ^{ - 1}}\left( 0 \right).
    Also we know, tan⁡(0)=0\tan \left( 0 \right) = 0.
    So, we can write it as,
    tan⁡−1(tan⁡0){\tan ^{ - 1}}\left( {\tan 0} \right)
    Now, using the property tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta , we get,
    tan⁡−1(tan⁡0)=0{\tan ^{ - 1}}\left( {\tan 0} \right) = 0
  4. tan⁡−1(2cos⁡(2π3)){\tan ^{ - 1}}\left( {2\cos \left( {\dfrac{{2\pi }}{3}} \right)} \right)
    The angle 2π3\dfrac{{2\pi }}{3} lies in the 2nd quadrant.
    So, we can write it as, 2π3=π−π3\dfrac{{2\pi }}{3} = \pi - \dfrac{\pi }{3}.
    Substituting this value in the cosine part, we get,
    cos⁡(π−π3)\cos \left( {\pi - \dfrac{\pi }{3}} \right)
    Since, cosine functions are negative in the 2nd quadrant, so,
    cos⁡2π3=cos⁡(π−π3)=−cos⁡π3\cos \dfrac{{2\pi }}{3} = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \dfrac{\pi }{3}
    ⇒cos⁡2π3=cos⁡(π−π3)=−12\Rightarrow \cos \dfrac{{2\pi }}{3} = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \dfrac{1}{2}
    Now, replacing this value in the inverse trigonometric function, we get,
    tan⁡−1(2(−12)){\tan ^{ - 1}}\left( {2\left( { - \dfrac{1}{2}} \right)} \right)
    =tan⁡−1(−1)= {\tan ^{ - 1}}\left( { - 1} \right)
    We know, the principle domain of tan⁡−1{\tan ^{ - 1}} lies in (−π2,π2)\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right).
    So, we can write,
    tan⁡(−π4)=−1\tan \left( {\dfrac{{ - \pi }}{4}} \right) = - 1
    So, we can write the inverse trigonometric function as,
    tan⁡−1(tan⁡(−π4)){\tan ^{ - 1}}\left( {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right)
    Now, using the property tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta , we get,
    tan⁡−1(tan⁡(−π4))=−π4{\tan ^{ - 1}}\left( {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right) = \dfrac{{ - \pi }}{4}

Note: The trigonometric functions are periodic circular functions and can extend within the Cartesian plane up to infinity. So, every trigonometric function as well as inverse trigonometric functions have their domains ( principle domains) already assigned, in case in the question no domain is specified, we are to find the value of the trigonometric function within their principal domain.