Answer
1 + $\frac{\sqrt 3}{2}$
or
$ \frac{ 2 + \sqrt 3 }{2}$
Work Step by Step
Given expression
= $( \sin 60^{\circ} + \cos 60^{\circ})^{2}$
= $ \sin^{2} 60^{\circ}$ + 2 $ \sin 60^{\circ} \cos 60^{\circ}$ + $ \cos^{2} 60^{\circ}$
[ on expanding using identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$]
= $(\frac{\sqrt 3}{2})^{2}$ + 2$ \times \frac{\sqrt 3}{2} \times\frac{1}{2}$ + $ (\frac{1}{2})^{2}$
= $ \frac{3}{4} + \frac{\sqrt 3}{2} + \frac{1}{4}$
= $ \frac{3}{4} + \frac{1}{4} + \frac{\sqrt 3}{2} $
= $ \frac{(3+1)}{4} + \frac{\sqrt 3}{2} $
= $ \frac{4}{4} + \frac{\sqrt 3}{2} $
= 1 +$ \frac{\sqrt 3}{2} $
or $\frac{2 + \sqrt 3}{2}$
