A) \[A>B\]
B) \[A=B\]
C) \[A<B\]
D) none of these
Correct Answer: A
Solution :
| \[A=2{{\tan }^{-1}}(2\sqrt{2}-1)>2ta{{n}^{-1}}\sqrt{3}=\frac{2\pi }{3}\] \[\Rightarrow A>\frac{2\pi }{3}.\,\,\,Now\frac{1}{3}<\frac{\sqrt{3}}{2}\] |
| \[={{\sin }^{-1}}(1/3)<si{{n}^{-1}}\left( \frac{\sqrt{3}}{2} \right)=\frac{\pi }{3};{{\sin }^{-1}}\left( \frac{3}{5} \right)<{{\sin }^{-1}}\left( \frac{\sqrt{3}}{2} \right)=\frac{\pi }{3}.\]\[B={{\sin }^{-1}}\left( \frac{1}{3} \right)+{{\sin }^{-1}}\left( \frac{3}{5} \right)<\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}\] |
| Now \[A>\frac{2\pi }{3},\,\,B<\frac{2\pi }{3}.\,\,\therefore A>B.\] |
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