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Question Number 34422 by math1967 last updated on 06/May/18

$$Showthat$$$$\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}=1whenx=\frac{\sqrt{3}}{2}$$

Answered by Rio Mike last updated on 06/May/18

$$\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$$$$\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$$$$\frac{2+\sqrt{3}}{2(1+\sqrt{1+\frac{\sqrt{3}}{2}})}+\frac{2-\sqrt{3}}{2(1-\sqrt{1-\sqrt{\frac{\sqrt{3}}{2}})})}$$$$\frac{(2+\sqrt{3})(1-\sqrt{1+\frac{\sqrt{3}}{2}})}{-\sqrt{3}}+2+2\sqrt{1-\frac{\sqrt{3}}{2}}-\sqrt{3}-\sqrt{3(1-\frac{\sqrt{3}}{2})}$$$$\frac{-2\sqrt{3}+2\sqrt{3+\frac{3\sqrt{3}}{2}}-3+\sqrt{9+\frac{9\sqrt{3}}{2}}+2\sqrt{3}+2\sqrt{3-\frac{3\sqrt{3}}{2}}-3-\sqrt{9}-\frac{9\sqrt{3}}{2}}{3}$$$$\frac{2\sqrt{3+\frac{3\sqrt{3}}{2}}-6+\sqrt{9+\frac{9\sqrt{3}}{2}}+2\sqrt{3-\frac{3\sqrt{3}}{2}}-\sqrt{9}-\frac{9\sqrt{3}}{2}}{3}$$$$\approx 1$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/May/18

$$firstfind\sqrt{1+x}and\sqrt{1-x}$$$$1+x=1+\frac{\sqrt{3}}{2}and1-x=1-\frac{\sqrt{3}}{2}$$$$=\frac{2+\sqrt{3}}{2}and1-x=\frac{2-\sqrt{3}}{2}$$$$=\frac{4+2\sqrt{3}}{4}and=\frac{4-2\sqrt{3}}{4}$$$$=\frac{{(\sqrt{3}+1)}^2}{2^2}and=\frac{{(\sqrt{3}-1)}^2}{2^2}$$$$so\sqrt{1+x}=\frac{\sqrt{3}+1}{2}and\sqrt{1-x}=\frac{\sqrt{3}-1}{2}$$$$=\frac{2+\sqrt{3}}{2(1+\frac{\sqrt{3}+1}{2})}+\frac{2-\sqrt{3}}{2(1-\frac{\sqrt{3}-1}{2})}$$$$=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}$$$$=\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{9-3}$$$$=6/6=1Ans$$$$$$

Commented by Rasheed.Sindhi last updated on 06/May/18

$$\mathcal{F}ine!$$

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