If the solve the problem

Given:

$y=(\sin 2 x+\cot x+2)^{2} a t x=\frac{\pi}{2}$

First, we have to find $\frac{d y}{d x}$ of given function, $f(x)$,i.e, to find the derivative of $f(x)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$

The Slope of the tangent is $\frac{d y}{d x}$

$\Rightarrow y=(\sin 2 x+\cot x+2)^{2}$

$\frac{d y}{d x}=2 x(\sin 2 x+\cot x+2)^{2}-1\left\{\frac{d y}{d x}(\sin 2 x)+\frac{d y}{d x}(\cot x)+\frac{d y}{d x}(2)\right\}$

$\Rightarrow \frac{d y}{d x}=2(\sin 2 x+\cot x+2)\left\{(\cos 2 x) \times 2+\left(-\operatorname{cosec}^{2} x\right)+(0)\right\}$

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}(\sin x)=\cos x$

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}(\cot x)=-\operatorname{cosec}^{2} x$

$\Rightarrow \frac{d y}{d x}=2(\sin 2 x+\cot x+2)\left(2 \cos 2 x-\operatorname{cosec}^{2} x\right)$

Since, $x=\frac{\pi}{2}$

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=2 \times\left(\sin 2\left(\frac{\pi}{2}\right)+\cot \left(\frac{\pi}{2}\right)+2\right)\left(2 \cos 2\left(\frac{\pi}{2}\right)-\operatorname{cosec}^{2}\left(\frac{\pi}{2}\right)\right)$

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=2 \times\left(\sin (\pi)+\cot \left(\frac{\pi}{2}\right)+2\right) \times\left(2 \cos (\pi)-\operatorname{cosec}^{2}\left(\frac{\pi}{2}\right)\right)$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=2 \times(0+0+2) \times(2(-1)-1)$

$\therefore \sin (\pi)=0, \cos (\pi)=-1$

$\therefore \cot \left(\frac{\pi}{2}\right)=0, \operatorname{cosec}\left(\frac{\pi}{2}\right)=1$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=2(2) \times(-2-1)$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=4 \times-3$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=-12$

$\therefore$ The Slope of the tangent at $x=\frac{\pi}{2}$ is $-12$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\text { The Slope of the tangent }}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{-12}$

$\Rightarrow$ The Slope of the normal $=\frac{1}{12}$