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 4: find the principal value of begin align tan 1....
Question 4

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

Answer

\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}

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