A) 1
B) \[2/\sqrt{3}\]
C) \[2-\sqrt{3}\]
D) 2
Correct Answer: B
Solution :
Given \[x=\frac{\sqrt{3}}{2}\] \[\frac{\sqrt{1+x}}{1+\sqrt{1+x}}\times \frac{1-\sqrt{1+x}}{1-\sqrt{1+x}}+\frac{\sqrt{1-x}}{1-\sqrt{1-x}}\times \frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}\] \[\frac{\sqrt{1+}x-1-x}{1-1-x}+\frac{\sqrt{1-}x+1-x}{1-1+x}\] \[=\frac{\sqrt{1-}x+1-x}{x}-\frac{\sqrt{1+}x-1-x}{x}\] \[=\frac{\sqrt{1-}x+1-x-\sqrt{1+}x+1+x}{x}\] \[=\frac{2+\sqrt{1-}x-\sqrt{1+x}}{x}\] \[=\frac{2+\sqrt{1-\frac{\sqrt{3}}{2}}-\sqrt{1+\frac{\sqrt{3}}{2}}}{\frac{\sqrt{3}}{2}}\] \[=\frac{2+\sqrt{\frac{2-\sqrt{3}}{2}}-\sqrt{2+\frac{\sqrt{3}}{2}}}{\frac{\sqrt{3}}{2}}\] \[=\frac{2+\frac{\sqrt{4-2\sqrt{3}}}{2}-\frac{\sqrt{4+2\sqrt{3}}}{2}}{\frac{\sqrt{3}}{2}}\] \[=\frac{4+\sqrt{3}-1-\sqrt{3}-1}{\sqrt{3}}=\frac{2}{\sqrt{3}}\]
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