Class 12 Mathematics - Chapter Inverse Trigonometric Functions NCERT Solutions | Find the principal value of \begin{align
Find the principal value of \begin{align} sin^{-1}\left(-\frac{1}{2}\right)\end{align}
\begin{align} sin^{-1}\left(-\frac{1}{2}\right)=y \;\;Then\;\; sin y = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align}
We know that the range of the principal value branch of sin−1 is
\begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and \;\;sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\end{align}
Therefore, the principal value of
\begin{align} sin^{-1}\left(-\frac{1}{2}\right) is -\frac{\pi}{6}\end{align}
More Questions From Class 12 Mathematics - Chapter Inverse Trigonometric Functions
- Q:- Find the principal value of \begin{align} tan^{-1}\left(-\sqrt3\right)\end{align}
- Q:- Find the principal value of \begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right)\end{align}
- Q:- Find the value of if \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align}
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\begin{align} (A) \pi \;\;(B) -\frac{\pi}{3}\;\; (C) \frac{\pi}{3} \;\;(D) \frac{2\pi}{3}\end{align} - Q:- Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right)\end{align}
- Q:- Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{2}\right)\end{align}
- Q:- Find the principal value of \begin{align} cos^{-1}\left(\frac{1}{2}\right) + 2sin^{-1}\left(\frac{1}{2}\right)\end{align}
- Q:- Find the principal value of \begin{align} cos^{-1}\left(\frac{\sqrt3}{2}\right)\end{align}
- Q:- Find the principal value of \begin{align} tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}
- Q:- Find the principal value of \begin{align} cosec^{-1}\left({-\sqrt2}\right)\end{align}
- Q:- Find the principal value of \begin{align} cot^{-1}\left(\sqrt3\right)\end{align}
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(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y}
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