Question

If $si{n}^{-1}\frac{1}{3}+si{n}^{-1}\frac{2}{3}=si{n}^{-1}x,$ then x is equal to
[Roorkee 1995]

• A. \N
• B. $\frac{\sqrt{5}-4\sqrt{2}}{9}$
• C. $\frac{\sqrt{5}+4\sqrt{2}}{9}$
• D. $\frac{x}{2}$

Answer 1

The correct option is C

$\frac{\sqrt{5}+4\sqrt{2}}{9}$

$si{n}^{-1}\frac{1}{3}+si{n}^{-1}\frac{2}{3}=si{n}^{-1}[\frac{1}{3}\sqrt{1-\frac{4}{9}}+\frac{2}{3}\sqrt{1-\frac{1}{9}}]=si{n}^{-1}\left[\frac{\sqrt{5}+4\sqrt{2}}{9}\right]$
Therefore $x=\frac{\sqrt{5}+4\sqrt{2}}{9}$