Question

If ` tan (pi/12 - x) , tan (pi/12) , tan (pi/12 + x) ` in G.P. then sum of all the solutions in [0,314] is `k pi`. Find k

Answer 1

`tan^2pi/12=tan(pi/12-x)*tan(pi/12+x)`
`tan^2pi/12=((tanpi/12+tanx)/(a-tanpi/12*tanx))*((tanpi/12-tanx)/(1+tanpi/12tanx))`
`tan^(2pi/12)=(tam^2pi/12-tan^2x)/(1-tan^2pi/12*tan^2x)`
`tan^2(pi/12)(1-tan^2pi/12*tan^2x)=tan^2pi/12-tan^2x`
`tan^2pi/12-tan^4pi/12tan^2x=tan^2pi/12-tan^2x`
`tan^2x-tan^4pi/12tan^2x=0`
`tan^2x[1-tan^4pi/12]=0`
`tan^2x=0`
`tanx=0`
`x=npi`
`=pi[1+2+3+..+99]`
`=pi(99*100)/2`
`=4950pi`
`=kpi`
`k=4950`.