QuestionReportMathematics Question on Trigonometric Functionscos(40?+θ)+cos(120?+θ)+cos(220?+θ)+cos(300?+θ)=\cos(40? + \theta) + \cos(120? + \theta) + \cos(220? + \theta) + \cos(300? + \theta) =cos(40?+θ)+cos(120?+θ)+cos(220?+θ)+cos(300?+θ)=A3B2C1D0Answer0ExplanationSolutionPut θ=0∘\theta = 0^\circθ=0∘