Let tk=tan ^2(kπ)/(7) for k=1,2,3. Find the value of (t1+t2+t3)+t1t2t3.

The numbers $\tan\frac{k\pi}{7}$ for $k=1,2,3$ together with their negatives are the roots of $\tan7\theta=0$, $\theta\neq0$. Writing $\tan7\theta=\frac{7T-35T^{3}+21T^{5}-T^{7}}{1-21T^{2}+35T^{4}-7T^{6}}$ and setting the numerator to zero (after dividing by $T$) gives $T^{6}-21T^{4}+35T^{2}-7=0$. Its roots in $u=T^{2}$ are exactly $t_1,t_2,t_3$ from $u^{3}-21u^{2}+35u-7=0$. By Vieta, $t_1+t_2+t_3=21$ and $t_1t_2t_3=7$. The required value is $21+7=28$. A direct numerical evaluation of $\tan^2(\pi/7)+\tan^2(2\pi/7)+\tan^2(3\pi/7)=21.0000$ and the product $=7.0000$ confirms this. The same idea works for any odd $n$: the squares of the tangents of the acute multiples of $\pi/n$ are roots of a polynomial in $u=T^{2}$, so symmetric functions follow from the coefficients without evaluating a single tangent. Here the constant term gives the product directly, which is why the final answer is a small integer.