February 2nd, 2015, 12:45 PM  #1 
Member Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10  How do you prove this?
Prove: (tan(1))^2 + (tan(3))^2+....(tan(89))^2 = 45x(901) = 4005

February 2nd, 2015, 01:00 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra 
You are presumably going to form pairs$$\tan^2 x + \tan^2{\left( \tfrac\pi2  x \right) }$$and see what that comes out as.

February 2nd, 2015, 01:36 PM  #3 
Member Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10 
I've tried pairing them like that: so each angle x except 45 (where tan^2(45)=1) forms a pair with 90x. tan^2(x)+tan^2(90x)=tan^2(x)+cot^2(x)=sin^2(x)/cos^2(x)+cos^2(x)/sin^2(x)= 1/(cos^2(x)sin^2(x)). Hmm.... not sure what to do from here Idk if this is the right way to proceed: 1/(sin(2x)/2)^2=4/sin^2(x) Now the sum comes down to 1+4(1/sin^2(1)+1/sin^2(3)+....1/sin^2(43)) No idea what to do from here idk if stuff telescopes or something. 
February 2nd, 2015, 01:58 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,162 Thanks: 1638 
What makes you think the original equation is correct?

February 2nd, 2015, 02:03 PM  #5 
Member Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10  
February 3rd, 2015, 04:05 PM  #6  
Global Moderator Joined: Dec 2006 Posts: 19,162 Thanks: 1638 
So you need an initial assumption, don't you? It's already been made implicitly. Quote:
Use the fact that 1/sin²(x) = csc²(x) = 1 + cot²(x). Now you can try pairing again, but you'll need to be resourceful, as the pairs don't seem to be present at first. This is a wellknown problem, but I can't find a website where the full details of the above process are given.  

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