A) \[\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{2},\frac{1}{2}\]
C) \[\frac{1}{2},-\frac{1}{2}\]
D) \[\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\]
Correct Answer: D
Solution :
[d] Given, \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=\frac{\pi }{2}\] and \[{{\cos }^{-1}}x-{{\cos }^{-1}}y=0\] \[\Rightarrow \left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)-\left( \frac{\pi }{2}-{{\sin }^{-1}}y \right)=0\] \[\Rightarrow {{\sin }^{-1}}y-{{\sin }^{-1}}x=0\Rightarrow {{\sin }^{-1}}y={{\sin }^{-1}}x\] 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) \[y=\frac{1}{\sqrt{2}}\]
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