Double Symmetrical

We have that, $\tan{a}=\dfrac{\tan{(n+1)a}-\tan{na}}{1+\tan{(n+1)a}\times \tan{an}}\\ \Longrightarrow 1+\tan{(n+1)a}\tan{na}=\dfrac{\tan{(n+1)a}-\tan{na}}{\tan{a}}\\ \Longrightarrow (1+\tan{a}\tan{2a})+(1+\tan{2a}\tan{3a})+...(1+\tan{8a}\tan{9a})\\= \dfrac{\tan{2a}-\tan{a}+\tan{3a}-\tan{2a}+...\tan{9a}-\tan{8a}}{\tan{a}}\\=\dfrac{\tan{9a}-\tan{a}}{\tan{a}}\\=\dfrac{-2 \tan{a}}{\tan{a}}=-2\\ \Longrightarrow \text{our answer is}-2-8=-10$ .And done!

We note that:

$\tan{\frac{\pi}{5}} = - \tan{\frac{4\pi}{5}} = \tan{\frac{6\pi}{5}} = - \tan{\frac{9\pi}{5}} \\ \tan{\frac{2\pi}{5}} = - \tan{\frac{3\pi}{5}} = \tan{\frac{7\pi}{5}} = -\tan {\frac{8\pi}{5}} \\ \tan{\frac{5\pi}{5}} = \tan{\pi} = 0$

Therefore,

$f(a) = \tan{\frac{\pi}{5}} \tan{\frac{2\pi}{5}} + \tan{\frac{2\pi}{5}} \tan{\frac{3\pi}{5}} + \tan{\frac{3\pi}{5}} \tan{\frac{4\pi}{5}} + ... \tan{\frac{8\pi}{5}} \tan{\frac{9\pi}{5}} \\ \quad \quad = \tan{\frac{\pi}{5}} \tan{\frac{2\pi}{5}} - \tan^2{\frac{2\pi}{5}} + \tan{\frac{\pi}{5}} \tan{\frac{2\pi}{5}} + 0 \\ \quad \quad \quad + 0 + \tan{\frac{\pi}{5}} \tan{\frac{2\pi}{5}} - \tan^2{\frac{2\pi}{5}} + \tan{\frac{\pi}{5}} \tan{\frac{2\pi}{5}} \\ \quad \quad = 4\tan{\frac{\pi}{5}} \tan{\frac{2\pi}{5}} - 2 \tan^2{\frac{2\pi}{5}}$

Since $\frac{\pi}{5} + \frac{2\pi}{5} + \frac{2\pi}{5} = \pi$ , there are three angles of a triangle.

$\Rightarrow \tan{\frac{\pi}{5}} \tan^2{\frac{2\pi}{5}} = \tan{\frac{\pi}{5}} + 2 \tan{\frac{2\pi}{5}} \\ \tan{\frac{\pi}{5}} \tan^2{\frac{2\pi}{5}} = \tan{\frac{\pi}{5}} + \dfrac {4\tan{\frac{\pi}{5}}} {1-\tan^2{\frac{\pi}{5}}} \\ \Rightarrow \tan^2 {\frac {2\pi}{5}} = 1 + \dfrac {4} {1-\tan^2{\frac{\pi}{5}}} = \dfrac {5 - \tan^2{\frac{\pi}{5}}} {1-\tan^2{\frac{\pi}{5}}}$

$\Rightarrow f(a) = 4\tan{\frac{\pi}{5}} \tan{\frac{2\pi}{5}} - 2 \tan^2{\frac{2\pi}{5}} \\ \quad \quad \quad \space = \dfrac {8\tan^2{\frac{\pi}{5}}} {1-\tan^2 {\frac{\pi}{5}}} - \dfrac {10 - 2\tan^2{\frac{\pi}{5}}} {1-\tan^2{\frac{\pi}{5}}} \\ \quad \quad \quad \space = \dfrac {10\tan^2{\frac{\pi}{5}} - 10} {1-\tan^2{\frac{\pi}{5}}} = \boxed{-10}$

Write tana as a difference of numbers like tan(2a-a) and expand it accordingly now yoiu will get the value if 1+ (tan2atana)and by this you can find values of other as tana is in the denominator all the other terms cancels out leaving only tan9a and tana then by some manipulations you'll get -10

Similar to the explanation of @Adarsh Kumar.

Note that $\tan{x}\tan{y}=\frac{\tan{x}-\tan{y}}{\tan(x-y)}-1$ Therefore, we have $\sum_{k=1}^{8}{\tan{k\alpha}\tan(k+1)\alpha} =\sum_{k=1}^{8}( \frac{\tan(k+1)\alpha-\tan{k\alpha}}{\tan{\alpha}}-1) =\frac{\tan{9\alpha -\tan\alpha}}{\tan{\alpha}}-8$ Since $10\alpha=2\pi$ , it holds that $\tan{9\alpha}=-\tan{\alpha}$ . Now we can get the answer: $\frac{\tan{9\alpha -\tan\alpha}}{\tan{\alpha}}-8=\frac{-2\tan\alpha}{\tan\alpha}-8=\boxed{-10}$