Consider (tan3x - tanx)/2
=1/2(( sin3x/cos3x)-(sinx/cosx))
=1/2((sin3xcosx-cos3xsinx)/cos3xcosx)
=(sin(3x-x)/2cosxcos3x)
=2sinxcosx/2cosxcos3x
=sinx/cos3x
Therefore we can say that sin3x/cos9x=(tan9x-tan3x)/2
sin9x/cos27x=(tan27x-tan9x)/2
adding the three
we get
(sinx/cos3x)+(sin3x/cos9x)+(sin9x/cos27x)=(tan27x-tanx)/2
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