Given function
#f(x)=\cot^2x-x^2#
Now, the slope of the tangent line #
frac{d}{dx}f(x)# is given as
#\frac{d}{dx}f(x)=\frac{d}{dx}(\cot^2x-x^2)#
#f'(x)=2\cot x(-\cosec^2 x)-2x#
#f'(x)=-2\cosec^2 x\cot x-2x#
Now, the slope of tangent at the point #x=\pi/3#
#f'(x=\pi/3)=-2\cosec^2(\pi/3)\cot(\pi/3)-2(\pi/3)#
#=-2(2/\sqrt3)^2(1/\sqrt3)-{2\pi}/3#
#=-8/{3\sqrt3}-{2\pi}/3#
Hence, the slope of normal at the same point #x=\pi/3# is
#=-1/(-8/{3\sqrt3}-{2\pi}/3)#
#={3\sqrt3}/{8+2\pi\sqrt3}#
Now, the y-coordinate of point #x=\pi/3# is given by substituting #x=\pi/3# in the given function,
#y=f(x=\pi/3)=\cot^2(\pi/3)-(\pi/3)^2#
#y=1/3-\pi^2/9#
Hence, the equation of normal to the given curve at #(\pi/3, 1/3-\pi^2/9)# & having slope #{3\sqrt3}/{8+2\pi\sqrt3}# is given as follows
#y-(1/3-\pi^2/9)={3\sqrt3}/{8+2\pi\sqrt3}(x-\pi/3)#
#y-1/3+\pi^2/9={3\sqrt3}/{8+2\pi\sqrt3}(x-\pi/3)#
