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If `x_1, x_2, x_3,a n dx_4`are the roots of the equations `x^4-x^3sin2beta+x^2cos2beta-xcosbeta-sinbeta=0,`prove that `tan^(-1)x_1+tan^(-1)x_2+tan^(-1)x_3+tan^(-1)x_4=npi+(pi/2)-beta`, where `n`is an integer. |
Answer» Given equation is, `x^4-x^3sin2beta+x^2cos2beta - xcosbeta - sinbeta = 0` As `x_1,x_2,x_3 and x_4` are the roots of the above equation. `:. sum x_1 = sin2beta` `sumx_1x_2 = cos2beta` `sumx_1x_2x_3 = cosbeta` `x_1x_2x_3x_4 = -sinbeta` Now, `tan(tan^-1x_1+tan^-1x_2+tan^-1x_3+tan^-1x_4) = (sumx_1-sum x_1x_2x_3)/(1-sumx_1x_2+x_1x_2x_3x_4)` `=(sin2beta-cosbeta)/(1+cos2beta-sinbeta)` `=(2sinbetacosbeta-cosbeta)/(2sin^2beta - sinbeta)` `=(cosbeta(2sinbeta-1))/(sinbeta(2sinbeta -1))` `=cot beta = tan(pi/2-beta)` `:. tan(tan^-1x_1+tan^-1x_2+tan^-1x_3+tan^-1x_4) = tan(pi/2-beta)` `:. tan^-1x_1+tan^-1x_2+tan^-1x_3+tan^-1x_4 = npi+pi/2-beta`, where `n in Z`. |
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