Please answer question no 12 in the attached file Share with your friends Share 0 Neha Sethi answered this Dear student a) sinθ2=12a^-b^We have:a^-b^2=a^-b^.a^-b^ as a→.a→=a→2⇒a^-b^2=a^.a^-a^.b^-b^.a^+b^.b^⇒a^-b^2=a^2-2a^.b^+b^2 as a^.b^=b^.a^⇒a^-b^2=a^2-2a^b^cosθ+b^2 as a^.b^=a^b^cosθ ⇒a^-b^2=2-2cosθ asa^= b^=1⇒a^-b^2=21-cosθ=22sin2θ2=4sin2θ2⇒sin2θ2=14a^-b^2⇒sinθ2=12a^-b^Hence proved. Regards 0 View Full Answer