#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}#
