MCQ
જો ${\sin ^{ - 1}}\frac{1}{3} + {\sin ^{ - 1}}\frac{2}{3} = {\sin ^{ - 1}}x,$ તો $x = . . ..$
- A$0$
- B$\frac{{\sqrt 5 - 4\sqrt 2 }}{9}$
- ✓$\frac{{\sqrt 5 + 4\sqrt 2 }}{9}$
- D$\frac{\pi }{2}$
$ = {\sin ^{ - 1}}\left[ {\frac{1}{3}\sqrt {1 - \frac{4}{9}} + \frac{2}{3}\sqrt {1 - \frac{1}{9}} } \right] = {\sin ^{ - 1}}\,\left[ {\frac{{\sqrt 5 + 4\sqrt 2 }}{9}} \right]$
Therefore $x = \frac{{\sqrt 5 + 4\sqrt 2 }}{9}$.
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