IF k1= tan27Θ-tanΘ and k2= sinΘ/cos3Θ + sin3Θ/cos9Θ + sin9Θ/cos27Θ then prove that k1=2k2

Taking sinx/cos3x..........where x=θ2sinxcosx/2cos3xcosx.....(multiplying and dividing by 2cosx).=sin2x/2cos3xcosx=sin(3x-x)/2cos3xcosx=1/2(tan3x-tanx)...Similarly other two .. .and we get K2=1/2[(tan3x-tanx)+(tan9x-tan3x)+(tan27x-tan9x)].K2=1/2(tan27x-tanx)K2=1/2(k1)•••••k1=2k2...my first attempt..on "`askiitians""Thankyou