Note that the given expression simplifies to $3 \, cot3x$

$cotx +\frac{{\cos (60 + x)}}{{\sin (60 + x)}}\,\, + \,\,\frac{{\cos (x – 60)}}{{\sin (x – 60)}}$

$=$ $\frac{{\cos x}}{{\sin x}}\,\, + \,\,\frac{{\sin (2x)}}{{\sin (x + 60)\,\sin (x – 60)}}$

$=$$\frac{{\cos x}}{{\sin x}}\,\, + \,\,\frac{{8\,\sin x\,\,\cos x}}{{4{{\sin }^2}x\, – \,3}}$ 

$=$ $\frac{{4\,{{\sin }^2}x\,\,\cos x\,\, – \,3\cos x\,\, + \,8\,{{\sin }^2}x\,\cos x}}{{4{{\sin }^3}x\, – \,3\,\sin x}}$

$=$ $\frac{{3[3\cos x – 4{{\cos }^3}x]}}{{{{\sin }^3}x}}\,\,$

$= 3\, cot3x$

$\Rightarrow$ $\frac{{3[1 – 3{{\tan }^2}x]}}{{3\tan x\, – \,{{\tan }^3}x}}\,\,$