Inverse Trigonometric Functions Question 188
Question: $ {{\sin }^{-1}}[ x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}} ]= $
Options:
A) $ {{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{x} $
B) $ {{\sin }^{-1}}x-{{\sin }^{-1}}\sqrt{x} $
C) $ {{\sin }^{-1}}\sqrt{x}-{{\sin }^{-1}}x $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ x=\sin \theta $ and $ \sqrt{x}=\sin \varphi $ Hence $ {{\sin }^{-1}}(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}}) $
$ ={{\sin }^{-1}}(\sin \theta \sqrt{1-{{\sin }^{2}}\varphi }-\sin \varphi \sqrt{1-{{\sin }^{2}}\theta }) $
$ ={{\sin }^{-1}}(\sin \theta \cos \varphi -\sin \varphi \cos \theta )={{\sin }^{-1}}\sin (\theta -\varphi ) $
$ =\theta -\varphi ={{\sin }^{-1}}(x)-{{\sin }^{-1}}(\sqrt{x}) $ .
BETA
