If a = \(\frac{\sqrt {3}}{2}\) then find the value of \(\sqrt {1 + a}\) + \(\sqrt {1 - a}\) + \(\frac{\sqrt {3}}{2}\)
Answer & explanation
Correct answer: option 2
a = \(\frac{\sqrt {3}}{2}\)
1 + a = 1 + \(\frac{\sqrt {3}}{2}\) = \(\frac{2 + \sqrt {3}}{2}\)
= \(\frac{4 + 2\sqrt {3}}{4}\) (multiply & divide by 2)
= \(\frac{(\sqrt {3} + 1)^2}{4}\)
Similarly,
1 - a = \(\frac{(\sqrt {3} - 1)^2}{4}\)
Now,
⇒ \(\sqrt {1 + a}\) + \(\sqrt {1 - a}\) + \(\frac{\sqrt {3}}{2}\) = \(\sqrt {(\frac{\sqrt {3} + 1)^2}{4}}\) + \(\sqrt {(\frac{\sqrt {3} - 1)^2}{4}}\) + \(\frac{\sqrt {3}}{2}\)
= \(\frac{\sqrt {3}\;+\;1}{2}\) + \(\frac{\sqrt {3}\;-\;1}{2}\) + \(\frac{\sqrt {3}}{2}\)
= \(\frac{\sqrt {3}\;+\;1\;+\sqrt {3}\;-\;1}{2}\) + \(\frac{\sqrt {3}}{2}\)
= \(\sqrt {3}\) + \(\frac{\sqrt {3}}{2}\)
= 3\(\frac{\sqrt {3}}{2}\)