Plot a 3D graph while changing the range of the parameter

Question

I wish to plot the graph of a surface while being able to change a parameter.

My current code is

ContourPlot3D[(Cos[Pi/4]+x Cos[0]Sin[Pi/4]+y Sin[0]Sin[Pi/4])^2 == 
 (1+x^2+y^2)Cos[a]^2, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}, 
 {a = 0 .. Pi}, AxesLabel -> {x, y, z}]

I don't want to do it with Manipulate[], since I don't a the graph of the surface of a given "a" between 0 and Pi, but the graph of the surface with "a" from 0 to Pi.

Answer 1

Your syntax is wrong, as you probably know. Maybe you want something like this:

ContourPlot3D[
 Evaluate@Table[
   (Cos[Pi/4] + x*Cos[0] Sin[Pi/4] + y*Sin[0] Sin[Pi/4])^2 == (1 + 
       x^2 + y^2) Cos[a]^2, {a, 0, \[Pi], .2}], {x, -10, 10}, {y, -10,
   10}, {z, -10, 10}, AxesLabel -> {x, y, z}]

Or

Show@Table[
  ContourPlot3D[
   (Cos[Pi/4] + x*Cos[0] Sin[Pi/4] + y*Sin[0] Sin[Pi/4])^2 == (1 + 
       x^2 + y^2) Cos[a]^2, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}, 
   Mesh \[Rule] None, AxesLabel -> {x, y, z}, 
   ContourStyle \[Rule] FaceForm[{Pink, Opacity[0.8]}]
   ],
  {a, 0, \[Pi], .5}
  ]

Answer 2

May be (I am doing a lot of guessing, but it seems consistent with @acl's guess!)

Table[ContourPlot[(Cos[Pi/4] + x Cos[0] Sin[Pi/4] + 
        y Sin[0] Sin[Pi/4])^2 == (1 + x^2 + y^2) Cos[a]^2 // 
    Release, {x, -10, 10}, {y, -10, 10}, AxesLabel -> {x, y, z},
   ContourStyle -> Hue[a/Pi // N]],
  {a, 0, Pi, Pi/16}] // Show