Find the angle of intersection of curves, $y=\left[\right|\sin x\left|\right|\cos x\left|\right]$ and $x^2{y}^2=5$, where [.] denotes the greatest integral function.

We know that, 

$1\le \left|\sin x\right|\left|\cos x\right|\le \sqrt{2}$

$\therefore y=\left[\left|\sin x\right|+\left|\cos x\right|\right]=1$

Let $P$ and $Q$ be the points of intersection of given curves.

Clearly, the given curves meet at points where $y=1$, so we get

 $x^2+1=5$

$$\begin{gathered} \Rightarrow x^2=4 \\ \Rightarrow x=\pm 2 \end{gathered}$$

Now, we have $P(2,1)$ and $Q(-2,1)$

Differentiating $x^2+y^2=5$ w.r.t. $x$, we get

$$\begin{gathered} 2x+2y\left(\frac{dy}{dx}\right)=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-x}{y} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,1)}=-2 \end{gathered}$$

and ${\left(\frac{dy}{dx}\right)}_{(-2,1)}=2$

Clearly, the slope of line $y=1$ is zero and the slope of the tangents at $P$ and $Q$ are $(-2)$ and $\left(2\right)$, respectively.

Thus, the angle of intersection is ${\tan }^{-1}\left(2\right)$.

Hence,  the correct option is $1.$