Inverse Trigonometric Functions Question 196
Question: If $ {{\sin }^{-1}}x+{{\sin }^{-1}}y=\pi /2 $ and $ {{\cos }^{-1}}x-{{\cos }^{-1}}y=0. $ then values x and y are respectively
Options:
A) $ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} $
B) $ \frac{1}{2},\frac{1}{2} $
C) $ \frac{1}{2},-\frac{1}{2} $
D) $ \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} $
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Answer:
Correct Answer: D
Given, $ {{\sin }^{-1}}x+{{\sin }^{-1}}y=\frac{\pi }{2} $ and $ {{\cos }^{-1}}x-{{\cos }^{-1}}y=0 $
$ \Rightarrow ( \frac{\pi }{2}-{{\sin }^{-1}}x )-( \frac{\pi }{2}-{{\sin }^{-1}}y )=0 $
$ \Rightarrow {{\sin }^{-1}}y-{{\sin }^{-1}}x=0\Rightarrow {{\sin }^{-1}}y={{\sin }^{-1}}x $
From equations (i) and (ii), we get $ 2{{\sin }^{-1}}x=\frac{\pi }{2} $
$ \Rightarrow {{\sin }^{-1}}x=\frac{\pi }{4}\Rightarrow x=\frac{1}{\sqrt{2}} $
From equation (ii) $ y=\frac{1}{\sqrt{2}} $
BETA
