| 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) \[\frac{8\pi }{3}\]
B) \[\frac{4\pi }{3}\]
C) \[\frac{2\pi }{3}\]
D) \[\frac{\pi }{3}\]
Correct Answer: D
Solution :
\[\because \] \[\cos \,\left( \frac{4\pi }{3} \right)=-\frac{1}{2}<0\] \[\therefore \] \[\frac{\pi }{2}<{{\cos }^{-1}}\left( \cos \,\frac{4\pi }{3} \right)\le \pi \] \[\therefore \] \[{{\cos }^{-1}}\,\cos \,\left( \frac{4\pi }{3} \right)\,=\frac{2\pi }{3}\] 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, \[{{\sin }^{-1}}\left( \sin \,\frac{4\pi }{3} \right)+{{\cos }^{-1}}\,\left( \cos \,\frac{4\pi }{3} \right)=\frac{\pi }{3}\]
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