$\displaystyle T_{2,0}(x) = f(0)+f'(0)x+\frac{f''(0)}{2}x^2 $
It's easy to see that $\displaystyle f(0)=0 $.
$\displaystyle f'(0)=\sqrt{\cos(0)} = 1 $
$\displaystyle f''(x) = \left(\sqrt{\cos(x)}\right)' = -\frac{\sin(x)}{2\sqrt{\cos(x)}} $, so $\displaystyle f''(0) = 0 $
Thus $\displaystyle T_{2,0}(x) = x $.