Find the principal value of egin{align | Class 12 Mathematics Chapter Inverse Trigonometric Functions, Inverse Trigonometric Functions NCERT Solutions

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

\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}

 


Study Tips for Answering NCERT Questions:

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:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Inverse Trigonometric Functions.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

Comments

Comment(s) on this Question

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}....