To find the slope, m, of the tangent line, we must compute #dy/dx# in terms of #theta# and then evaluate it at #theta = (3pi)/8#.

Here is a reference Tangents with Polar Coordinates that will give us the general equation for #dy/dx# in terms of #theta#:

#dy/dx = ((dr)/(d theta)sin(theta)+rcos(theta))/((dr)/(d theta)cos(theta)-rsin(theta))" [1]"#

We are given r:

#r(theta) = theta + cot(theta)+thetasin^3(theta)" [2]"#

We must compute #(dr)/(d theta)#

#(dr)/(d theta) = 1 - csc^2(θ) + sin^3(θ) + 3 θ sin^2(θ) cos(θ)" [3]"#

Substitute the right sides of equations [2] and [3] into equation [1]:

#dy/dx = ((1 - csc^2(θ) + sin^3(θ) + 3 θ sin^2(θ) cos(θ))sin(theta)+(theta + cot(theta)+thetasin^3(theta))cos(theta))/((1 - csc^2(θ) + sin^3(θ) + 3 θ sin^2(θ) cos(θ))cos(theta)-(theta + cot(theta)+thetasin^3(theta))sin(theta))" [4]"#

When we evaluate equation [4] at #theta = (3pi)/8#, the left side of the equation becomes the slope, m:

#m = ((1 - csc^2((3pi)/8) + sin^3((3pi)/8) + 3 (3pi)/8 sin^2((3pi)/8) cos((3pi)/8))sin((3pi)/8)+((3pi)/8 + cot((3pi)/8)+(3pi)/8sin^3((3pi)/8))cos((3pi)/8))/((1 - csc^2((3pi)/8) + sin^3((3pi)/8) + 3 (3pi)/8 sin^2((3pi)/8) cos((3pi)/8))cos((3pi)/8)-((3pi)/8 + cot((3pi)/8)+(3pi)/8sin^3((3pi)/8))sin((3pi)/8))" [5]"#

#m ~~ -1.58#