Inverse Trigonometric Functions Question 24
Question: If $ {{\sin }^{-1}}\frac{1}{3}+{{\sin }^{-1}}\frac{2}{3}={{\sin }^{-1}}x, $ then x is equal to
[Roorkee 1995]
Options:
A) 0
B) $ \frac{\sqrt{5}-4\sqrt{2}}{9} $
C) $ \frac{\sqrt{5}+4\sqrt{2}}{9} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: C
Solution:
$ {{\sin }^{-1}}\frac{1}{3}+{{\sin }^{-1}}\frac{2}{3} $
$ ={{\sin }^{-1}}[ \frac{1}{3}\sqrt{1-\frac{4}{9}}+\frac{2}{3}\sqrt{1-\frac{1}{9}} ]={{\sin }^{-1}}[ \frac{\sqrt{5}+4\sqrt{2}}{9} ] $
Therefore $ x=\frac{\sqrt{5}+4\sqrt{2}}{9} $ .
BETA
