Inverse Trigonometric Functions Question Answers: NCERT Class 12 Mathematics

\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} 

\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} 

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

\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}

\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} 

\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}

\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}

\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}

\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} 

\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}

\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}

\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}

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}