| 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, | |
A) 0
B) \[2\pi -10\]
C) \[-\pi \]
D) \[3\pi -10\]
Correct Answer: C
Solution :
\[\because \] \[5=5\times {{57}^{o}}\approx {{285}^{o}}\] \[\therefore \] \[\sin \,5<0\] and \[\cos \,5\,<0\] \[\Rightarrow \] \[{{\sin }^{-1}}\,(\sin \,5)\,={{\sin }^{-1}}\,\{\sin \,(\pi -5)\}\,=\pi -5\] and \[{{\cos }^{-1}}\,(\cos \,5)=co{{s}^{-1}}\,\{\cos \,(2\pi -5)\}\] \[=2\pi -5\] \[\therefore \] \[{{\sin }^{-1}}(\sin \,5)\,-{{\cos }^{-1}}\,(\cos \,5)\] \[=\pi -5-2\pi +5=-\pi \]
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