If x cos θ=y cos (θ+2π3)=z cos(θ+4π3), then the value of 1x+1y+1z is equal to
We have
x cos θ=y cos (θ+2π3)=z cos(θ+4π3)=k⇒cos θ=kx, cos(θ+2π3)=kyand cos(θ+4π3)=kzHence kx+ky+kz=cos θ+cos(θ+2π3)+cos(θ+4π3)
=cos θ+cos(π3−θ)−cos(π3+θ)=cos θ−2 cosπ3cos θ=0