Class 12 Mathematics - Chapter Inverse Trigonometric Functions NCERT Solutions | Find the principal value of \begin{align

Welcome to the NCERT Solutions for Class 12th Mathematics - Chapter Inverse Trigonometric Functions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 8: find the principal value of begin align cot 1....
Question 8

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

Answer

\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}

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