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 10: find the principal value of begin align cosec....
Question 10

Find the principal value of \begin{align} cosec^{-1}\left({-\sqrt2}\right)\end{align}

Answer

\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}

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