If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ be given by $\mathrm{f}\left(\mathrm{x}\right)={\sin }^2 \mathrm{x}+{\sin }^2\left(x+\frac{\mathrm{\pi }}{3}\right)+\cos \mathrm{xcos} \left(\mathrm{x}+\frac{\mathrm{\pi }}{3}\right)$ for all x$\in$R and g : R$\to$R be such that$\mathrm{g}\left(\frac{5}{4}\right)=1$, then $g\circ f\left(x\right)$ is ___________

The given function is $f\left(x\right)={\sin }^{2}x+{\sin }^2\left(x+\frac{\mathrm{\pi }}{3}\right)+\cos x\cos \left(x+\frac{\mathrm{\pi }}{3}\right)$

This can be written as 

$\begin{array}{rcl}f\left(x\right)& =& \frac{1-\cos 2x}{2}+\frac{1-\cos \left(2x+{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{2}+\frac{1}{2}\left(2\cos x\cos \left(x+\frac{\mathrm{\pi }}{3}\right)\right)\\ & =& \frac{1}{2}\left(2-\cos 2x-\cos \left(2x+\frac{2\mathrm{\pi }}{3}\right)+\cos \frac{\mathrm{\pi }}{3}+\cos \left(2x+\frac{\mathrm{\pi }}{3}\right)\right)\\ & =& \frac{1}{2}\left[\frac{5}{2}-\cos 2x+\cos \left(\frac{\mathrm{\pi }}{3}-2x\right)+\cos \left(\frac{\mathrm{\pi }}{3}-2x\right)\right]\\ & =& \frac{1}{2}\left(\frac{5}{2}-\cos 2x+\cos 2x\right)\\ & =& \frac{5}{4}\end{array}$

Here the function $f\left(x\right)$ is constant function and given $g\left(\frac{5}{4}\right)=1$

Therefore, $g\circ f\left(x\right)=1$