how to calculate the exact value of $\tan \frac{\pi}{10}$
Your textbook probably has an example, where $\cos(\pi/5)$ (or $\sin(\pi/5)$) has been worked out. I betcha it also has formulas for $\sin(\alpha/2)$ and $\cos(\alpha/2)$ expressed in terms of $\cos\alpha$. Take it from there.
look this How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$?
then you will get $\sin \frac{\pi}{10}$($\frac{2\pi }{5}+\frac{\pi}{10}=\frac{\pi}{2}$) ,then $\tan \frac{\pi}{10}$.
$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ \cos\pars{\bracks{n + 1}\theta} + \cos\pars{\bracks{n - 1}\theta} = 2\cos\pars{n\theta}\cos\pars{\theta} $$ $$ \cos\pars{\bracks{n + 1}\theta} = 2\cos\pars{n\theta}\cos\pars{\theta} - \cos\pars{\bracks{n - 1}\theta} $$ Let's $\theta = \pi/10$ and $x = \cos\pars{\theta}$: \begin{align} \cos\pars{2\theta} &= 2x^{2} - 1 \tag{1} \\ \cos\pars{3\theta} &= \bracks{2\cos\pars{2\theta} - 1}x \tag{2} \\ \cos\pars{4\theta} &= 2\cos\pars{3\theta}x - \cos\pars{2\theta} \tag{3} \\ 0 &= 2\cos\pars{4\theta}x - \cos\pars{3\theta} \tag{4} \end{align}
$\pars{3}$ and $\pars{4}$ yield: $$ 0 = \pars{4x^{2} - 1}\cos\pars{3\theta} - 2\cos\pars{2\theta}x \tag{5} $$
$\pars{2}$ and $\pars{5}$ yield: $$ 0 = \pars{4x^{2} - 1}\bracks{2\cos\pars{2\theta} - 1}x - 2\cos\pars{2\theta}x = \pars{8x^{3} - 4x}\cos\pars{2\theta} - 4x^{3} + x $$ $$ 4\pars{2x^{2} - 1}\cos\pars{2\theta} - 4x^{2} + 1 = 0 \tag{6} $$
$\pars{1}$ and $\pars{6}$ yield: $$ 0 = 4\pars{2x^{2} - 1}^{2} - 4x^{2} + 1 = 4\pars{2x^{2} - 1}^{2} - 2\pars{2x^{2} - 1} - 1 $$
Then, $$ \pars{2x^{2} - 1}_{\pm} = {2 \pm \sqrt{\pars{-2}^{2} - 4\times 4\times\pars{-1}} \over 2\times 4} = {1 \pm \sqrt{5} \over 4} $$ Obviously, we take the "$+$ sign" as a solution: $$ 2x^{2} - 1 = {1 + \sqrt{5} \over 4} \quad\imp\quad x = \cos\pars{\pi \over 10} = \sqrt{{1 \over 2}\bracks{1 + {1 + \sqrt{5} \over 4}}} = \sqrt{{5 + \sqrt{5} \over 8}} $$ Then, \begin{align} \tan\pars{\pi \over 10} &= \sqrt{\bracks{1 \over \cos\pars{\pi/10}}^{2} - 1} = \sqrt{{8 \over 5 + \sqrt{5\,}} - 1} = \sqrt{3 - \sqrt{5\,} \over 5 + \sqrt{5\,}} = \sqrt{1 - 2\,{1 + \sqrt{5\,} \over 5 + \sqrt{5}}} \\[3mm]&= \sqrt{1 - 2\,{\pars{1 + \sqrt{5\,}}\pars{5 - \sqrt{5\,}} \over 20}} = \sqrt{1 - {4\sqrt{5\,} \over 10}} = \sqrt{1 - {2 \over \sqrt{5\,}}} \\[5mm]& \end{align}
$${\large% \tan\pars{\pi \over 10} = \sqrt{1 - {2 \over \sqrt{5\,}}}} $$