Solve
 $$\tan(A + B + C) = \dfrac{\sum \tan A \prod \tan A}{1 - \sum \tan A \tan B}$$

Solution

$$\tan \left(\dfrac{2 \pi}{5} - \dfrac{\pi}{15}\right)$$
$$\tan (A + B + C) = \dfrac{\displaystyle \sum \tan A . \pi \tan A}{1 - \displaystyle \sum \tan A. \tan B}$$
$$\tan (A + B + C) = \dfrac{(\tan A + \tan B + \tan C) \tan A. \tan B . \tan C}{1 - (\tan A. \tan B + \tan B + \tan C + \tan C . \tan A)}$$
$$\dfrac{(\tan A + \tan B + \tan C) - \tan A . \tan B . \tan C}{1 - (\tan A . \tan B + \tan C. \tan A + \tan B . \tan C)}$$ = 
Both are equal Hence proved