Number of ordered pair  $(a,b)$ for each of which the equality  $a(\cos x-1)+b^2=\cos (ax+b^2)-1$ holds true  $\forall x\in R$ are

Put  $x=0\Rightarrow b^2=\cos b^2-1\Rightarrow \cos b^2=1+b^2$

$\Rightarrow b=0$

For  $b=0$ we have  $a(\cos x-1)=\cos ax-1$

$\Rightarrow 2a{\sin }^2\frac{x}{2}=2{\sin }^2\frac{ax}{2}$

$\Rightarrow a=0$  or  $a=1$

Hence ordered pairs are  $(0,0),(1,0)$.