Question

If x = √3/2, then show that 1+x/1+root1+x + 1-x/1-root1-x = 1​

Answer 1

i dont no this question answer

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

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

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

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

$\bf \underline{Solution-}$

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

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

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

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

$\textsf{Similarly,}$

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

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

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

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

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

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

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

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

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

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

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

$\rm{= 1}$

$\textsf{LHS = RHS}$

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