\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}
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 7: Find the principal value of egin{align} sec^{-1}left(frac{2}{sqrt3} ight)end{align}....
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