Prove

Prove that :-

\sum_{r=1}^{n}{tan\ r\ \alpha }\ .\ tan\ (r+1)\ \alpha = cot\ \alpha \ . \ tan\ (n+1)\ \alpha\ - n - 1

2 Answers

106

Asish Mahapatra ·2010-07-08 05:01:33

tan[(r+1)@ - r@] = tan(r+1)@ - tan(r@)1+tan(r+1)@tan(r@)

=> tan(r@) tan(r+1)@ = tan(r+1)@ - tan(r@)tan@ - 1

now u can sum this. to get tan(n+1)@ - tan@tan@ - n which is equivalnet to the answer asked

correct [1]

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