A) \[\alpha >\beta \]
B) \[\alpha =\beta \]
C) \[\alpha <\beta \]
D) \[\alpha +\beta =2\pi \]
Correct Answer: C
Solution :
Now,\[\alpha +\beta ={{\sin }^{-1}}\frac{\sqrt{3}}{2}+{{\cos }^{-1}}\frac{\sqrt{3}}{2}\] \[+{{\sin }^{-1}}\frac{1}{3}+{{\cos }^{-1}}\frac{1}{3}\] \[=\frac{\pi }{2}+\frac{\pi }{2}=\pi \] Also, \[\alpha =\frac{\pi }{3}+{{\sin }^{-1}}\frac{1}{3}<\frac{\pi }{3}+{{\sin }^{-1}}\frac{1}{2}\] As\[\sin \theta \]is increasing in\[\left[ 0,\,\,\frac{\pi }{2} \right]\]. \[\therefore \] \[\alpha <\frac{\pi }{3}+\frac{\pi }{6}=\frac{\pi }{2}\] \[\Rightarrow \] \[\beta >\frac{\pi }{2}>\alpha \] \[\Rightarrow \] \[\alpha <\beta \]
You need to login to perform this action.
You will be redirected in
3 sec
