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Math Help - summation question

  1. #1
    MHF Contributor
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    summation question

    Simplify

    \sum^{1995}_{n=1}\tan n\tan (n+1)
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  2. #2
    Super Member
    Joined
    Jan 2009
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    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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