Inverse Trigonometric Functions Question 162
Question: If $ x $ takes non-positive permissible value, then $ {{\sin }^{-1}}x $ =
Options:
A) $ {{\cos }^{-1}}\sqrt{1-x^{2}} $
B) $ -{{\cos }^{-1}}\sqrt{1-x^{2}} $
C) $ {{\cos }^{-1}}\sqrt{x^{2}-1} $
D) $ \pi -{{\cos }^{-1}}\sqrt{1-x^{2}} $
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Answer:
Correct Answer: B
Solution:
Let $ {{\sin }^{-1}}x=y. $ Then $ x=\sin y $
Since $ -1\le x\le 0, $
Therefore $ \frac{-\pi }{2}\le {{\sin }^{-1}}x\le 0 $
and so $ \frac{-\pi }{2}\le y\le 0 $ We have $ \cos y=\sqrt{1-{{\sin }^{2}}y} $
$ \Rightarrow \cos y=\sqrt{1-x^{2}} $ , for $ 0\le y\le \pi $ -. .(i)
Now $ -\frac{\pi }{2}\le y\le 0\Rightarrow \frac{\pi }{2}\ge -y\ge 0 $
$ \Rightarrow \cos ( -y )=\sqrt{1-x^{2}} $ {from (i)}
$ \Rightarrow -y={{\cos }^{-1}}\sqrt{1-x^{2}}\Rightarrow y=-{{\cos }^{-1}}\sqrt{1-x^{2}} $ .
BETA
