\(sin^{-1}(\frac{-1}{2})+2cos^{-1}(\frac{-\sqrt{3}}{2})=?\) A. \(\fr

1.	\(sin^{-1}(\frac{-1}{2})+2cos^{-1}(\frac{-\sqrt{3}}{2})=?\) A. \(\frac{\pi}{2}\) B. \({\pi}\)C. \(\frac{3\pi}{2}\)D. none of these
Answer» Correct Answer is (C) \(\frac{3\pi}{2}\) Let , x = \(sin^{-1}(\frac{-1}{2})+2cos^{-1}(\frac{-\sqrt{3}}{2})\) ⇒ \(sin^{-1}(\frac{1}{2})+2[\pi -cos^{-1}(\frac{-\sqrt{3}}{2})]\) ( \(\because\) sin-1(-\(\theta\)) = -sin^-1(\(\theta\)) and cos^-1(-\(\theta\))=π - cos^-1(\(\theta\))] ⇒ x= \(-(\frac{\pi}{6})\)+2\([\pi-\frac{\pi}{6}]\) ⇒ x= \(-(\frac{\pi}{6})\)+\(2[\frac{5\pi}{6}]\) ⇒ x = \(-\frac{\pi}{6}\)\(+\frac{5\pi}{3}\) ⇒ x = \(\frac{3\pi}{2}\)

\(sin^{-1}(\frac{-1}{2})+2cos^{-1}(\frac{-\sqrt{3}}{2})=?\) A. \(\frac{\pi}{2}\) B. \({\pi}\)C. \(\frac{3\pi}{2}\)D. none of these

Answer»

Correct Answer is (C) \(\frac{3\pi}{2}\)

Let , x = \(sin^{-1}(\frac{-1}{2})+2cos^{-1}(\frac{-\sqrt{3}}{2})\)

⇒ \(sin^{-1}(\frac{1}{2})+2[\pi -cos^{-1}(\frac{-\sqrt{3}}{2})]\)

( \(\because\) sin-1(-\(\theta\)) = -sin^-1(\(\theta\)) and cos^-1(-\(\theta\))=π - cos^-1(\(\theta\))]

⇒ x= \(-(\frac{\pi}{6})\)+2\([\pi-\frac{\pi}{6}]\)

⇒ x= \(-(\frac{\pi}{6})\)+\(2[\frac{5\pi}{6}]\)

⇒ x = \(-\frac{\pi}{6}\)\(+\frac{5\pi}{3}\)

⇒ x = \(\frac{3\pi}{2}\)