Simplify
$\displaystyle \sum^{1995}_{n=1}\tan n\tan (n+1)$
Use the formula $\displaystyle \tan(A-B) = \frac{ \tan(A) - \tan(B) }{ 1 + \tan(A) \tan(B) } $ , sub $\displaystyle A = n+1 ~,~ B=n $ we have
$\displaystyle \tan(1) = \frac{ \tan(n+1) - \tan(n) }{ 1 + \tan(n) \tan(n+1) } $ or
$\displaystyle \tan(n) \tan(n+1) = \frac{ \tan(n+1) - \tan(n) }{ \tan(1) } - 1 $ .
Therefore ,
$\displaystyle \sum_{n=1}^{m} \tan(n) \tan(n+1) = \frac{\tan(m+1) - \tan(1)}{\tan(1)} -m = \frac{\tan(m+1) }{\tan(1)} - ( m+1) $