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)
