Question

Somebody please solve this question. wrong answers will be reported​

Answer 1

$Given \: a = \frac{\sqrt{3}}{2}$

$i) \sqrt{ 1 + a }$

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

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

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

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

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

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

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

/* Similarly */

$ii) \sqrt{ 1 - a }$

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

$Now, LHS = \sqrt{ 1 + a } + \sqrt{ 1 - a }$

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

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

$= \frac{2}{2}$

$= 1$

$= RHS$

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Answer 2

Answer:

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