#f(x) = cot^2 x + sin(x-pi/2) #

First, find the slope of the tangent line of #f(x)# at #x=(5pi)/6# by finding the derivative of #f(x)#:

#f'(x) = 2(cotx)(-csc^2 x) + cos(x-pi/2)#

#f'(x) = frac{-2cosx}{sin^3 x} + cos(x-pi/2)#

#f'((5pi)/6) = frac{-2cos((5pi)/6)}{sin^3 ((5pi)/6)} + cos(frac{5pi-3pi}{6})#

# = frac{-2(-sqrt3/2)}{(1/2)^3} + cos(pi/3)#

# = frac{sqrt3}{1/8} + 1/2#

# = 8sqrt3 + 1/2#

Use common denominators to combine the two terms into one:

# f'((5pi)/6)= frac{16sqrt3 + 1}{2}#

Now that we have the slope of the tangent line at the point where #x=(5pi)/6#, find the slope of the normal line at this same point by taking the opposite reciprocal of this value:

#"Normal line slope" = frac{-1}{frac{16sqrt3 + 1}{2}}#

#"Normal line slope" = frac{-2}{16sqrt3 + 1}#