Question

please answer my question no.10

Answer 1

HELLO DEAR,



R.H.S

$\begin{lgathered}\frac{1}{1 + \sqrt{3} + \sqrt{5} } \\ \\ = > \frac{1(1 + \sqrt{3}- \sqrt{5} ) }{(1 + \sqrt{3} - 5)(1 + \sqrt{3} + \sqrt{5}) } \\ \\ = > \frac{1 + \sqrt{3} - \sqrt{5} }{ {(1 + \sqrt{3} )}^{2} - {( \sqrt{5} )}^{2} } \\ \\ = > \frac{1 + \sqrt{3} - \sqrt{5} }{1 + 3 + 2 \sqrt{3} - 5 } \\ \\ = > \frac{(1 + \sqrt{3} - \sqrt{5} ) }{(2 \sqrt{3} - 1)}\end{lgathered}$


now multipy both nuumerator and denominator by [2√3 +1]



$\begin{lgathered}\frac{(1 + \sqrt{3} - \sqrt{5} )(2 \sqrt{3} + 1 )}{(2 \sqrt{3 } - 1)(2 \sqrt{3} + 1) } \\ \\ = > \frac{2 \sqrt{3} + 1 + 6 - 2 \sqrt{15} - \sqrt{5} }{12 - 1} \\ \\ = > \frac{7 + 2 \sqrt{3} - \sqrt{5} - 2 \sqrt{15} }{11} \\ \\ = > \frac{7}{11} + \frac{2 \sqrt{3} }{11} - \frac{ \sqrt{5} }{11} - \frac{2 \sqrt{15} }{11}\end{lgathered}$

ON COMPARING WITH L.H.S,

we get,

p = 7/11


I HOPE ITS HELP YOU DEAR,
THANKS