If cos⁡xa=cos⁡(x+θ)b=cos⁡(x+2θ)c=cos⁡(x+3θ)d then a+cb+d is equal to

For each of the ratio s be 1/k, a+cb+d=kcos⁡x+kcos⁡(x+2θ)kcos⁡(x+θ)+kcos⁡(x+3θ)=2cos⁡(x+θ)cos⁡θ2cos⁡(x+2θ)cos⁡θ=cos⁡(x+θ)cos⁡(x+2θ)=kcos⁡(x+θ)kcos⁡(x+2θ)=bc