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Given $A_ 1 , A_2,A_3,...A_n$, and if $A_i=A $ for $i=1,2,...,n$ prove that

$$\tan(A_1)\tan(A_2)+\tan(A_1)\tan(A_3)+$$ ...+ i.e sum of tangents taken two at a time is ${n}\choose{2}$$\tan^2{A}$

Now i can see that binomial coefficient comes because we are taking two tangents at a time from n But i have no idea how to prove this

THANKS!!!

Sophie Clad
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    See http://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series and http://math.stackexchange.com/questions/951522/trig-sum-tan-21-circ-tan-22-circ-tan2-89-circ – lab bhattacharjee Feb 09 '17 at 05:48
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    What's relation among $A_ 1 , A_2,A_3,...A_n$.? – Nosrati Feb 09 '17 at 05:48
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    You are asking something about $A$ which is nowhere defined in the question. – dxiv Feb 09 '17 at 05:53
  • all A's are angle – Sophie Clad Feb 09 '17 at 05:54
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    @SophieClad What is $A$ in ${n}\choose{2}$$\tan^2{A},$? And what are the other $A_n,$, for that matter. Are they all related in some way that your question doesn't mention? – dxiv Feb 09 '17 at 05:56
  • @dxiv oh sorry i edited question – Sophie Clad Feb 09 '17 at 05:59
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    @SophieClad After your edit: since all $\tan$ terms are now equal, you can cancel them all out. Are you asking why $\sum_{i,j=1; i \lt j}^n 1 =\binom{n}{2},$? – dxiv Feb 09 '17 at 06:02
  • @dxiv i need to see how tan came to picture also – Sophie Clad Feb 09 '17 at 06:04
  • @SophieClad Sorry, I don't follow. You are saying that $A_i=A$ so $\tan(A_i) ,\tan(A_j) = \tan^2(A),$, then all $\tan$ terms cancel out between the two sides. You are then left to prove that the sum on the left-hand side has in fact $\binom{n}{2}$ terms. Maybe you should review and rethink your question. – dxiv Feb 09 '17 at 06:06
  • @dxiv yes i need to show how binomial coefficient came in there – Sophie Clad Feb 09 '17 at 06:29
  • @SophieClad Just count how many distinct terms you have in the sum on the left hand side. This is a basic combinatorics question, and has nothing to do with trigonometry. – dxiv Feb 09 '17 at 06:39

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