We know,
1−cosθ=2sin2(2θ)1+cosθ=2cos2(2θ)
Now,(a+b)⋅(a+b)=∣a∣2+2a⋅b+∣b∣2⇒(a+b)⋅(a+b)=(1)2+2×1×1×cosθ+(1)2⇒(a+b)⋅(a+b)=2(1+cosθ)⇒∣a+b∣2=4cos2(2θ)⇒∣a+b∣=2cos(2θ)...(1)
Similarly,
(a−b)⋅(a−b)=∣a∣2−2a⋅b+∣b∣2⇒(a−b)⋅(a−b)=(1)2−2×1×1×cosθ+(1)2⇒(a−b)⋅(a−b)=2(1−cosθ)⇒∣a−b∣2=4sin2(2θ)⇒∣a−b∣=2sin(2θ)...(2)
Therefore,
∣A^−B^∣=∣A^+B^∣2tanθ