From the reference Tangents with Polar Coordinates we obtain the equation
#dy/dx = ((dr)/(d theta)sin(theta)+rcos(theta))/((dr)/(d theta)cos(theta)-rsin(theta))" [1]"#
We are given #r = f(theta) = theta^2 - sec(theta)#; substitute this into equation [1]:
#dy/dx = ((dr)/(d theta)sin(theta)+(theta^2 - sec(theta))cos(theta))/((dr)/(d theta)cos(theta)-(theta^2 - sec(theta))sin(theta))" [2]"#
Compute #(dr)/(d theta)#:
#(dr)/(d theta) = 2theta-tan(theta)sec(theta)#
Substitute this into equation [2]:
#dy/dx = ((2theta-tan(theta)sec(theta))sin(theta)+(theta^2 - sec(theta))cos(theta))/((2theta-tan(theta)sec(theta))cos(theta)-(theta^2 - sec(theta))sin(theta))" [3]"#
The slope, m, of the tangent line at #theta = (3pi)/4# is the above equation evaluated at #theta = (3pi)/4#
#m = ((2(3pi)/4-tan((3pi)/4)sec((3pi)/4))sin((3pi)/4)+(((3pi)/4)^2 - sec((3pi)/4))cos((3pi)/4))/((2(3pi)/4-tan((3pi)/4)sec((3pi)/4))cos((3pi)/4)-(((3pi)/4)^2 - sec((3pi)/4))sin((3pi)/4))" [4]"#
#m = (9pi^2+32sqrt(2)-24pi)/(24pi + 9pi^2)#
The slope #m ~~ 0.357#
