2. Circumscribedarea = n Tan[Pi/n]

Code and Figures

Input := 

n1 = Sqrt[1 + Tan[Pi/5]^2];
n3 = Sqrt[1 + Tan[3 Pi/5]^2];
n5 = Sqrt[1 + Tan[5 Pi/5]^2];
n7 = Sqrt[1 + Tan[7 Pi/5]^2];

Input := 

c1 = 1; d1 = Tan[Pi/5];
c2 = -n1/n3; d2 = c2 Tan[3 Pi/5];
c3 = -n1/n5; d3 = c3 Tan[5 Pi/5];
c4 = -n1/n7; d4 = c4 Tan[7 Pi/5];
c5 = 1; d5 = Tan[9 Pi/5];

Input := 

lines = ParametricPlot[{{c1 t, d1 t}, {c2 t, d2 t},
   {c3 t, d3 t}, {c4 t, d4 t}, {c5 t, d5 t}}, {t, 0, 1},
   DisplayFunction -> Identity];

Input := 

tangents = ParametricPlot[{
 {c1 + (c2 - c1) t, d1 + (d2 - d1) t},
 {c2 + (c3 - c2) t, d2 + (d3 - d2) t},
 {c3 + (c4 - c3) t, d3 + (d4 - d3) t},
 {c4 + (c5 - c4) t, d4 + (d5 - d4) t},
 {c5 + (c1 - c5) t, d5 + (d1 - d5) t}},  {t, 0, 1},
 DisplayFunction -> Identity];

Input := 

p1 = Show[circle, lines, tangents,
 DisplayFunction -> $DisplayFunction,
 Prolog -> AbsoluteThickness[3]]

Output =

-Graphics-

Input := 

triangle = ParametricPlot[{{t, 0},
 {1, d1 t},
 {c1 t, d1 t}},
 {t, 0, 1},
 PlotStyle -> AbsoluteThickness[6],
 DisplayFunction -> Identity];

Input := 

Show[p1, triangle,
 DisplayFunction -> $DisplayFunction]

Output =

-Graphics-

If we call the angle from the x-axis inside the bold triangle theta = 2 Pi/n, then the area of the triangle is
(1/2) (1) Tan[theta]
So, the area of each of the n triangles formed by circumscribing the circle is
2 (1/2) (1) Tan[2 Pi/n] = Tan[2 Pi/n]
and since there are n such triangles, the area is
Area = n Tan[2 Pi/n].