The number of solutions of $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$ is

If $a, b, c$ are positive real numbers and the value of $\theta = \tan^{-1}\sqrt{\frac{a(a+b+c)}{bc}} + \tan^{-1}\sqrt{\frac{b(a+b+c)}{ca}} + \tan^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$,then $\tan \theta$ is equal to

If $p$ and $q$ are roots of $6x^2 + 10x + 1 = 0$,then the value of $[\tan^{-1} p + \tan^{-1} q]$ is: {where $[x]$ denotes the greatest integer less than or equal to $x$}

Consider the statements:
$(I)$ If $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(\tan ^{-1} x\right)\right)\right)$, then $f(0) = \frac{1}{2}$.
$(II)$ $\sin \left(4 \tan ^{-1} \frac{1}{5} - \tan ^{-1} \frac{1}{239}\right) = 1$.
Then the correct option among the following is:

The number of real solutions of $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ is

If $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$, then $x \in$