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2015-04-05T00:57:01+05:30

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Tan (a+b) =  √3      =>  a+b = π/3
tan (a-b)  =  1          =>  a-b = π/4

2a = 7 π / 12
10 a = 35 π/12  = 2 π + 11 π/12
tan 10 a =    tan (11 π/12)    = tan (π - π/12)
     =  - tan π/12
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another way:
   tan 2a =  tan (a+b + a -b) = [tan (a+b) + tan (a-b) ] / [ 1 - tan (a+b) tan (a-b) ]
           = [ √3 + 1 ] / [ 1 - √3 ]  = - (√3+1)² /2 = - (2 +√3)

  tan 4a = 2 tan 2a / [ 1 - tan² 2a ] = - 2 (2 +√3)/ [ 1 - 4 - 3 - 4√3 ]
           =  (2 +√3) / [√3 (2 + √3) ] =  1/√3

  tan 8a = (2/√3) / [ 1 - 1/3 ] = √3

  tan 10 a =  tan (8a +2a) = [ √3 - 2 -√3 ] / [ 1 + √3(2+√3) ]
               = - 1 / (2 + √3) = - (2 - √3)

2 5 2