Find the value of tan 18

Let θ = 18°. Then

5θ = 90°

∴ 2θ + 3θ = 90°

⇒ 2θ = 90° – 30°

⇒ sin 2θ = sin (90° – 3θ) = cos 3θ

⇒ 2 sin θ cos θ = 4 cos³θ – 3 cos θ

⇒ 2 sin θ cos θ – 4 cos³θ + 3 cos θ = 0

⇒ cos θ (2 sin θ – 4 cos² θ + 3) = 0

⇒ 2 sin θ – 4 (1 – sin²θ) + 3 = 0                            (cos θ = cos 18° ≠ 0)

⇒ 4sin² θ + 2 sin θ – 1 = 0

$$\begin{gathered} \mathrm{Now}, \\ \cos 18° = \sqrt{1 - {\sin }^{2}18°} \\ =\sqrt{1 - \frac{{\left(\sqrt{5}-1\right)}^2}{16}} \\ =\sqrt{\frac{16 - \left(5 + 1 - 2\sqrt{5}\right)}{16}} \\ =\sqrt{\frac{16-\left(6-2\sqrt{5}\right)}{16}} \\ =\sqrt{\frac{16 - 6 + 2\sqrt{5}}{16}} \\ =\sqrt{\frac{10+2\sqrt{5}}{16}} \\ =\frac{\sqrt{10+2\sqrt{5}}}{4} \\ \mathrm{Now}, \tan 18° = \frac{\sin 18°}{\cos 18°} = \frac{\sqrt{5} - 1}{4} \times \frac{4}{\sqrt{10+2\sqrt{5}}}=\frac{\sqrt{5} - 1}{\sqrt{10+2\sqrt{5}}} \end{gathered}$$