summation question

**mathaddict** · October 2nd 2010, 05:35 AM

Simplify

$\sum^{1995}_{n=1}\tan n\tan (n+1)$

**simplependulum** · October 2nd 2010, 06:53 AM

Use the formula $\tan(A-B) = \frac{ \tan(A) - \tan(B) }{ 1 + \tan(A) \tan(B) }$ , sub $A = n+1 ~,~ B=n$ we have

$\tan(1) = \frac{ \tan(n+1) - \tan(n) }{ 1 + \tan(n) \tan(n+1) }$ or

$\tan(n) \tan(n+1) = \frac{ \tan(n+1) - \tan(n) }{ \tan(1) } - 1$ .

Therefore ,

$\sum_{n=1}^{m} \tan(n) \tan(n+1) = \frac{\tan(m+1) - \tan(1)}{\tan(1)} -m = \frac{\tan(m+1) }{\tan(1)} - ( m+1)$

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