• # question_answer Direction: We have, $x<0$ $x\ge 0$ $\frac{-\pi }{2}\le {{\sin }^{-1}}x<0$ $0\le \,{{\sin }^{-1}}x\le \frac{\pi }{2}$ $\frac{\pi }{2}<{{\cos }^{-1}}x\le \pi$ $0\le {{\cos }^{-1}}x\le \frac{\pi }{2}$ Then, The principal value of ${{\sin }^{-1}}\left( \sin \,\frac{4\pi }{3} \right)+{{\cos }^{-1}}\cos \left( \frac{4\pi }{3} \right)$ is A)  $\frac{8\pi }{3}$                               B)  $\frac{4\pi }{3}$ C)  $\frac{2\pi }{3}$                               D)  $\frac{\pi }{3}$

$\because$ $\cos \,\left( \frac{4\pi }{3} \right)=-\frac{1}{2}<0$ $\therefore$    $\frac{\pi }{2}<{{\cos }^{-1}}\left( \cos \,\frac{4\pi }{3} \right)\le \pi$ $\therefore$    ${{\cos }^{-1}}\,\cos \,\left( \frac{4\pi }{3} \right)\,=\frac{2\pi }{3}$ and      $\sin \,\frac{4\pi }{3}=\frac{-\sqrt{3}}{2}<0$ $\therefore$    $-\frac{\pi }{2}\le {{\sin }^{-1}}\cdot \,\left( \sin \,\frac{4\pi }{3} \right)<0$ $\therefore$    ${{\sin }^{-1}}\left( \sin \,\frac{4\pi }{3} \right)=\frac{-\pi }{3}$ Hence, ${{\sin }^{-1}}\left( \sin \,\frac{4\pi }{3} \right)+{{\cos }^{-1}}\,\left( \cos \,\frac{4\pi }{3} \right)=\frac{\pi }{3}$