1. Given equation

We need the number of real roots of

Let

Then the equation is $A+B=\frac{\pi}{4}.$

2. Domain conditions

For the equation to be defined, both inverse trigonometric expressions must exist.

(i) For $\tan^{-1}\sqrt{x(x+1)}$

We need $x(x+1)\ge 0.$ This gives $x\in (-\infty,-1]\cup[0,\infty).$

(ii) For $\sin^{-1}\sqrt{x^2+x+1}$

Since $\sin^{-1}(t)$ is defined for $t\in[-1,1]$, and here $t=\sqrt{x^2+x+1}\ge 0$, we need $0\le \sqrt{x^2+x+1}\le 1.$ So, $x^2+x+1\le 1$ which gives $x^2+x\le 0$ $x(x+1)\le 0.$ Hence, $x\in[-1,0].$

3. Combine domain restrictions

From the first condition: $x\in (-\infty,-1]\cup[0,\infty)$ From the second condition: $x\in[-1,0].$

Their intersection is only $x=-1 \quad \text{or} \quad x=0.$

So only these two values can possibly satisfy the equation.

4. Check $x=-1$

Compute: $x(x+1)=(-1)(0)=0 \implies \sqrt{x(x+1)}=0$ so $\tan^{-1}(0)=0.$

Also, $x^2+x+1=1-1+1=1 \implies \sqrt{x^2+x+1}=1$ so $\sin^{-1}(1)=\frac{\pi}{2}.$

Thus LHS is $0+\frac{\pi}{2}=\frac{\pi}{2}\ne \frac{\pi}{4}.$ So $x=-1$ is not a root.

5. Check $x=0$

Compute: $x(x+1)=0 \implies \sqrt{x(x+1)}=0$ so $\tan^{-1}(0)=0.$

Also, $x^2+x+1=1 \implies \sqrt{x^2+x+1}=1$ so $\sin^{-1}(1)=\frac{\pi}{2}.$

Thus LHS is $0+\frac{\pi}{2}=\frac{\pi}{2}\ne \frac{\pi}{4}.$ So $x=0$ is not a root.

6. Conclusion

No real value of $x$ satisfies the equation.

Therefore, the number of real roots is $\boxed{0}.$

So the correct option is D.