Here is a reference to Tangents with polar coordinates
From the reference, we obtain the following equation:
dy/dx = ((dr)/(d theta)sin(theta) + rcos(theta))/((dr)/(d theta)cos(theta) - rsin(theta))
We need to compute (dr)/(d theta) but please observe that r(theta) can be simplified by using the identity sin(x)/cos(x) = tan(x):
r = -tan^2(theta)/theta
(dr)/(d theta) =(g(theta)/(h(theta)))' = (g'(theta)h(theta) - h'(theta)g(theta))/(h(theta))^2
g(theta) = -tan^2(theta)
g'(theta) = -2tan(theta)sec^2(theta)
h(theta) = theta
h'(theta) = 1
(dr)/(d theta) = (-2thetatan(theta)sec^2(theta) + tan^2(theta))/(theta)^2
Let's evaluate the above at pi/4
sec^2(pi/4) = 2
tan(pi/4) = 1
r'(pi/4) = (-2(pi/4)(1)(2) + 1)/(pi/4)^2
r'(pi/4) = (-2(pi/4)(1)(2) + 1)(16/(pi^2))
r'(pi/4) = (16- 16pi)/(pi^2)
Evaluate r at pi/4:
r(pi/4) = -4/pi = -(4pi)/pi^2
Note: I made the above denominator pi^2 so that it was common with the denominator of r' and would, therefore, cancel when we put them in the following equation:
dy/dx = ((dr)/(d theta)sin(theta) + rcos(theta))/((dr)/(d theta)cos(theta) - rsin(theta))
At pi/4 the sines and cosines are equal, therefore, they will cancel.
We are ready to write an equation for the slope, m:
m = (16 - 16pi + -4pi)/(16 - 16pi - -4pi)
m = (4 - 5pi)/(4 - 3pi)
