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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 |
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Answer» `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`. |
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