A) \[-\,\,\sqrt{3}\]
B) \[-\,\,1\]
C) \[1\]
D) \[\sqrt{3}\]
Correct Answer: D
Solution :
| \[x=\frac{\sqrt{3}}{2}\]\[\Rightarrow \]\[\frac{1}{x}=\frac{2}{\sqrt{3}}\] |
| By componendo and dividendo, |
| \[\frac{1+x}{1-x}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{{{(2+\sqrt{3})}^{2}}}{(2-\sqrt{3})(2+\sqrt{3})}=\frac{{{(2+\sqrt{3})}^{2}}}{4-3}\] |
| \[\Rightarrow \]\[\frac{1+x}{1-x}={{(2+\sqrt{3})}^{2}}\]\[\therefore \]\[\frac{\sqrt{1+x}}{\sqrt{1-x}}=\frac{2+\sqrt{3}}{1}\] |
| Again, by componendo and dividendo, |
| \[\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{2+\sqrt{3}+1}{2+\sqrt{3}-1}\] |
| \[=\frac{3+\sqrt{3}}{\sqrt{3}+1}=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1}=-\,\,\sqrt{3}\] |
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