Question

$\mathsf{If \: x = \frac{ \sqrt{3} }{2} , \: then \: show \: that \: \frac{1 + x}{1 + \sqrt{1 + x} } + \frac{1 - x}{1 - \sqrt{1 - x} } = 1}$

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Answer 1

$\huge\boxed{\boxed{\mathcal\pink{Good \: Morning \: }}}$

$\huge\mathcal\colorbox{lavender}{{\color{b}{✿Yøur-Añswer♡}}}$

$\large\bf{\underline{\red{illusion}}}$

$Given: x = </p><p>\frac{\sqrt{3}}{2}$

$To find: </p><p>\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}$

$\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}$

Put Value of x in fiven expresion we get,

$\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$

$\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}$

$\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{4+2\sqrt{3}}{4}}}+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{4-2\sqrt{3}}{4}}}$

$\frac{\frac{2+\sqrt{3}}{2}}{\frac{2+\sqrt{4+2\sqrt{3}}}{2}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2-\sqrt{4-2\sqrt{3}}}{2}}$

$\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}$

$\frac{(2+\sqrt{3})(2-\sqrt{4-2\sqrt{3}})+(2-\sqrt{3})(2+\sqrt{4+2\sqrt{3}})}{(2+\sqrt{4+2\sqrt{3}})(2-\sqrt{4-2\sqrt{3}})}$

$\frac{4-\sqrt{12-6\sqrt{3}}+2\sqrt{3}2\sqrt{4-2\sqrt{3}}+42\sqrt{3}-\sqrt{12+6\sqrt{3}}+2\sqrt{4+6\sqrt{3}}}{4+2\sqrt{4+2\sqrt{3}}-2\sqrt{4-2\sqrt{3}}-\sqrt{16-12}}$

$\frac{8-\sqrt{12-6\sqrt{3}}-\sqrt{12+6\sqrt{3}}+2\sqrt{4+2\sqrt{3}}-2\sqrt{4-2\sqrt{3}}}{2+2\sqrt{4+2\sqrt{3}}-2\sqrt{4-2\sqrt{3}}}$

$\frac{8-(3-3)-(3+\sqrt{3})+2(1+\sqrt{3})-2(1-\sqrt{3})}{2+2(1+\sqrt{3})-2(1-\sqrt{3})}$

$\frac{8-6+\sqrt{3}-\sqrt{3}+2-2+2\sqrt{3}+2\sqrt{3}}{2+2+2\sqrt{3}-2+2\sqrt{3}}$

$\frac{2+2\sqrt{3}+2\sqrt{3}}{2+2\sqrt{3}+2\sqrt{3}}$

⇒  1

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\rm{x = \frac{ \sqrt{3} }{2} }$

$\bf \underline{To \: show-}$

$\mathsf{ \frac{1 + x}{1 + \sqrt{1 + x} } + \frac{1 - x}{1 - \sqrt{1 - x} } = 1}$

$\bf \underline{Solution-}$

$\rm{ \sqrt{1 + x} } \\ = \sqrt{1 + \frac{ \sqrt{3} }{2} } \\ = \sqrt{ \frac{ 2 + \sqrt{3} }{2} } \\ = \sqrt{ \frac{4 + 2 \sqrt{3} }{4} }$

$= \sqrt{ \frac{3 + 1 + 2 \times \sqrt{3} \times 1}{4} }$

$= \sqrt{ \frac{( \sqrt{3} + 1 {)}^{2} }{4} }$

$= \frac{ \sqrt{3} + 1 }{2}$

$\textsf{Similarly,}$

$\rm{ \sqrt{1 - x} }$

$\rm{ = \frac{ \sqrt{3} - 1}{2} }$

$\rm{\therefore \: \frac{1 + x}{1 + \sqrt{1 + x} } + \frac{1 - x}{1 - \sqrt{1 - x} } }$

$\rm{= \frac{1 + \frac{ \sqrt{3} }{2} }{1 + \frac{ \sqrt{3} + 1 }{2} } + \frac{ 1 - \frac{ \sqrt{3} }{2} }{1 - \bigg( \frac{ \sqrt{3} - 1 }{2} \bigg) } }$

$\rm{= \frac{2 + \sqrt{3} }{2 + \sqrt{3} + 1 } + \frac{2 - \sqrt{3} }{2 - ( \sqrt{3} - 1)} }$

$\rm{= \frac{2 + \sqrt{3} }{3 + \sqrt{3} } + \frac{2 - \sqrt{3} }{3 - \sqrt{3} } }$

$\rm{= \frac{2 + \sqrt{3} }{3 + \sqrt{3} } \times \frac{3 - \sqrt{3} }{3 - \sqrt{3} } + \frac{2 - \sqrt{3} }{3 - \sqrt{3} } \times \frac{3 + \sqrt{3} }{3 + \sqrt{3} } }$

$\rm{= \frac{(2 + \sqrt{3} )(3 - \sqrt{3} ) + (2 - \sqrt{3})(3 + \sqrt{3} )}{(3 {)}^{2} - ( \sqrt{3} {)}^{2} } }$

$\rm{= \frac{6 - 2 \sqrt{3} + 3 \sqrt{3} - 3 + 6 + 2 \sqrt{3} - 3 \sqrt{3} - 3}{9 - 3} }$

$\rm{= \frac{12 - 6}{6} }$

$\rm{= \frac{6}{6} }$

$\rm{= 1}$

$\textsf{LHS = RHS}$

$\bf \underline{Hence \: proved.-}$