• # 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 ${{\sin }^{-1}}(\sin 5)-{{\cos }^{-1}}(\cos 5)$ is A)  0                                 B)  $2\pi -10$ C)  $-\pi$                        D)  $3\pi -10$

$\because$    $5=5\times {{57}^{o}}\approx {{285}^{o}}$ $\therefore$    $\sin \,5<0$ and $\cos \,5\,<0$ $\Rightarrow$ ${{\sin }^{-1}}\,(\sin \,5)\,={{\sin }^{-1}}\,\{\sin \,(\pi -5)\}\,=\pi -5$ and ${{\cos }^{-1}}\,(\cos \,5)=co{{s}^{-1}}\,\{\cos \,(2\pi -5)\}$ $=2\pi -5$ $\therefore$    ${{\sin }^{-1}}(\sin \,5)\,-{{\cos }^{-1}}\,(\cos \,5)$ $=\pi -5-2\pi +5=-\pi$