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Inverse Trigonometric Functions

In the previous chapter, types of functions are mentioned. If the function is not one-one, onto or both, then its inverse does not exist. This is the primary condition for inverse which must be fulfilled. Knowledge of the domain of range which we acquired earlier would help us in this chapter for graphing of inverse trigonometric functions. Trigonometric functions which are not one-one, onto or both will not be discussed. It has wide applications in engineering and other science related branches. This chapter consists of range, domain and principle value branches, graphs, elementary properties.

Download pdf of NCERT Solutions for Class mathematics Chapter 2 Inverse Trigonometric Functions

Download pdf of NCERT Examplar with Solutions for Class mathematics Chapter 2 Inverse Trigonometric Functions

Exercise 1

  • Q1 Find the principal value of \begin{align} sin^{-1}\left(-\frac{1}{2}\right)\end{align}
    Ans:

    \begin{align} sin^{-1}\left(-\frac{1}{2}\right)=y \;\;Then\;\; sin y = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align} 

    We know that the range of the principal value branch of sin−1 is

    \begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and \;\;sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\end{align} 

    Therefore, the principal value of

    \begin{align} sin^{-1}\left(-\frac{1}{2}\right) is -\frac{\pi}{6}\end{align} 


    Q2 Find the principal value of \begin{align} cos^{-1}\left(\frac{\sqrt3}{2}\right)\end{align}
    Ans:

    \begin{align} Let\;\; cos^{-1}\left(\frac{\sqrt3}{2}\right)=y, \;\;Then,\;\; cos y = \frac{\sqrt3}{2} = cos\left(\frac{\pi}{6}\right)\end{align} 

    We know that the range of the principal value branch of cos−1 is

     \begin{align} \left[0,\pi\right] and \;\;cos\left(\frac{\pi}{6}\right) = \frac{\sqrt3}{2}\end{align} 

    Therefore, the principal value of

     \begin{align} cos^{-1}\left(\frac{\sqrt3}{2}\right) is \frac{\pi}{6}\end{align} 


    Q3 Find the principal value of \begin{align} cosec^{-1}\left({2}\right)\end{align}
    Ans:

    \begin{align} Let \;\; cosec^{-1}\left({2}\right)=y \;\;Then\;\; cosec y = 2 = cosec\left(\frac{\pi}{6}\right)\end{align} 

    We know that the range of the principal value branch of cosec−1 is
    \begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \left(0 \right). \end{align}
    Therefore, the principal value of
    \begin{align} \frac{\pi}{6}\end{align}
     
     
     

    Q4 Find the principal value of \begin{align} tan^{-1}\left(-\sqrt3\right)\end{align}
    Ans:

    \begin{align} Let \;\; tan^{-1}\left(-\sqrt3\right)=y \;\;Then\;\; tan y = -\sqrt3 = -tan\left(\frac{\pi}{3}\right)= tan\left(-\frac{\pi}{3}\right)\end{align}

    We know that the range of the principal value branch of tan−1 is 

    \begin{align} \left(-\frac{\pi}{2},\frac{\pi}{2}\right) and \;\;tan\left(-\frac{\pi}{3}\right) = -\sqrt3\end{align}

     
    Therefore, the principal value of
     
    \begin{align} tan^{-1}\left(-\sqrt 3\right) is -\frac{\pi}{3}\end{align}

    Q5 Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{2}\right)\end{align}
    Ans:

    \begin{align} Let\;\; cos^{-1}\left(-\frac{1}{2}\right)=y, \;\;Then,\;\; cos y = -\frac{1}{2} = - cos\left(\frac{\pi}{3}\right)=cos\left(\pi - \frac{\pi}{3}\right) = cos\left(\frac{2\pi}{3}\right)\end{align}

    We know that the range of the principal value branch of cos−1 is 

     \begin{align} \left[0,\pi\right] and \;\;cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\end{align}

    Therefore, the principal value of

     \begin{align} cos^{-1}\left(-\frac{1}{2}\right) is \frac{2\pi}{3}\end{align} 


    Q6 Find the principal value of \begin{align} tan^{-1}\left(-1\right)\end{align}
    Ans:

    \begin{align} Let \;\; tan^{-1}\left(-1\right)=y. \;\;Then,\;\; tan y = -1 = -tan\left(\frac{\pi}{4}\right)= tan\left(-\frac{\pi}{4}\right)\end{align}

    We know that the range of the principal value branch of tan−1 is 

    \begin{align} \left(-\frac{\pi}{2},\frac{\pi}{2}\right) and \;\;tan\left(-\frac{\pi}{4}\right) = - 1\end{align}

    Therefore, the principal value of

    \begin{align} tan^{-1}\left(- 1\right) is -\frac{\pi}{4}\end{align}

     


    Q7 Find the principal value of \begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right)\end{align}
    Ans:

    \begin{align} Let \;\; sec^{-1}\left(\frac{2}{\sqrt3}\right)=y \;\;Then\;\; sec y = \frac{2}{\sqrt3} = sec\left(\frac{\pi}{6}\right)\end{align}

    We know that the range of the principal value branch of sec−1 is 

     \begin{align} \left[0,\pi\right] - \left(\frac{\pi}{2}\right) and \;\;sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt3}\end{align}

    Therefore, the principal value of

     \begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right) is \frac{\pi}{6}\end{align}

     


    Q8 Find the principal value of \begin{align} cot^{-1}\left(\sqrt3\right)\end{align}
    Ans:

    \begin{align} Let \;\; cot^{-1}\left(\sqrt3\right)=y. \;\;Then,\;\; cot y = \sqrt3 = cot\left(\frac{\pi}{6}\right).\end{align}

    We know that the range of the principal value branch of cot−1 is 

    \begin{align} \left(0, \pi\right) and \;\;cot\left(\frac{\pi}{6}\right) = \sqrt3.\end{align}

     
    Therefore, the principal value of
    \begin{align} cot^{-1}\left(\sqrt 3\right) is \frac{\pi}{6}\end{align}

    Q9 Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right)\end{align}
    Ans:

    \begin{align} Let\;\; cos^{-1}\left(-\frac{1}{\sqrt2}\right)=y, \;\;Then,\;\; cos y = -\frac{1}{\sqrt2} = - cos\left(\frac{\pi}{4}\right)=cos\left(\pi - \frac{\pi}{4}\right) = cos\left(\frac{3\pi}{4}\right)\end{align}

    We know that the range of the principal value branch of cos−1 is 

     \begin{align} \left[0,\pi\right] and \;\;cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt2}\end{align}

    Therefore, the principal value of

     \begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right) is \frac{3\pi}{4}\end{align} 


    Q10 Find the principal value of \begin{align} cosec^{-1}\left({-\sqrt2}\right)\end{align}
    Ans:

    \begin{align} Let \;\; cosec^{-1}\left({-\sqrt2}\right)=y \;\;Then\;\; cosec y = -{\sqrt2} =- cosec\left(\frac{\pi}{4}\right) = cosec\left(-\frac{\pi}{4}\right)\end{align} 

    We know that the range of the principal value branch of cosec−1 is

    \begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \left(0 \right) and \;\;cosec\left(-\frac{\pi}{4}\right) = -\sqrt2.\end{align}

    Therefore, the principal value of 

    \begin{align} cosec^{-1}\left(-\sqrt2\right) is -\frac{\pi}{4}\end{align}


    Q11 Find the principal value of \begin{align} tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}
    Ans:

    \begin{align} Let \;\; tan^{-1}(1)=x. \;\;Then\;\; tan x = 1 = tan\left(\frac{\pi}{4}\right).\end{align}

     \begin{align}  \therefore tan^{-1}(1)=tan\left(\frac{\pi}{4}\right)\end{align}

    \begin{align} Let \;\;cos^{-1}\left(-\frac{1}{2}\right)=y. \;\;Then,\;\; cos y = -\frac{1}{2} = -cos\left(\frac{\pi}{3}\right)= cos\left(\pi - \frac{\pi}{3}\right) = cos\left(\frac{2\pi}{3}\right)\end{align}

     \begin{align}  \therefore cos^{-1}\left(-\frac{1}{2}\right)  = \frac{2\pi}{3}\end{align}

    \begin{align} Let \;\; sin^{-1}\left(-\frac{1}{2}\right)=z. \;\;Then,\;\; sin z = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align}

     \begin{align}  \therefore sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\end{align}

     \begin{align}  \therefore tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}

    \begin{align} =\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}\end{align}

    \begin{align} =\frac{3\pi + 8\pi -2\pi}{12}=\frac{9\pi}{12}=\frac{3\pi}{4}\end{align}


    Q12 Find the principal value of \begin{align} cos^{-1}\left(\frac{1}{2}\right) + 2sin^{-1}\left(\frac{1}{2}\right)\end{align}
    Ans:

    \begin{align} Let \;\;cos^{-1}\left(\frac{1}{2}\right) =x. \;\;Then,\;\; cos x = \frac{1}{2} = cos\left(\frac{\pi}{3}\right).\end{align}

    \begin{align}  \therefore cos^{-1}\left(\frac{1}{2}\right)  = \frac{\pi}{3}\end{align}

    \begin{align} Let \;\; sin^{-1}\left(\frac{1}{2}\right)=y. \;\;Then,\;\; sin y = \frac{1}{2} = sin\left(\frac{\pi}{6}\right).\end{align}

    \begin{align}  \therefore sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\end{align}

    \begin{align} \therefore cos^{-1}\left(\frac{1}{2}\right) + 2sin^{-1}\left(\frac{1}{2}\right)\end{align}

    \begin{align} =\frac{\pi}{3} + \frac{2\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}\end{align}


    Q13 Find the value of if \begin{align} sin^{-1} x= y,\end{align}
    then
    \begin{align} (A) 0 ≤ y ≤ x \;\;\;\;\; (B) -\frac{\pi}{2} ≤ y ≤ \frac{\pi}{2} \end{align}
    \begin{align} (C)\; 0 < y < \pi \;\;\;\;\; (D) -\frac{\pi}{2} < y < \frac{\pi}{2} \end{align}
    Ans:

    It is given that sin−1 x = y.

    We know that the range of the principal value branch of sin−1 is 

    \begin{align}  \left[-\frac{\pi}{2} ,\frac{\pi}{2}\right] \end{align} 

    Therefore,

     \begin{align}  (B) -\frac{\pi}{2} ≤ y ≤ \frac{\pi}{2} \end{align}


    Q14 Find the value of if \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align}
    is equal to
    \begin{align} (A) \pi \;\;(B) -\frac{\pi}{3}\;\; (C) \frac{\pi}{3} \;\;(D) \frac{2\pi}{3}\end{align}
    Ans:
    \begin{align} Let \;\; tan^{-1}(\sqrt 3)=x. \;\;Then\;\; tan x = \sqrt 3 = tan\left(\frac{\pi}{3}\right).\end{align}
    We know that the range of the principal value branch of tan−1 is
      \begin{align} \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\end{align}
      \begin{align}  \therefore tan^{-1}\sqrt3=\frac{\pi}{3}\end{align}
    \begin{align} Let \;\; sec^{-1}\left(-2\right)=y \;\;Then\;\; sec y = -2 = -sec\left(\frac{\pi}{3}\right) = sec\left(\pi - \frac{\pi}{3}\right) = sec\left(\frac{2\pi}{3}\right)\end{align}
    We know that the range of the principal value branch of sec−1 is 
      \begin{align} \left[0,\pi\right] - \left(\frac{\pi}{2}\right)\end{align}
      \begin{align}  \therefore sec^{-1}({-2})=\frac{2\pi}{3}\end{align}
    Hence, \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align} 
    \begin{align} =\frac{\pi}{3} - \frac{2\pi}{3}=-\frac{\pi}{3}\end{align} 
     

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