Which of the following relations is true for two unit vector ^ A...

We know,

$1 - cos\theta = 2\sin^{2}\left(\dfrac{\theta}{2}\right)1 + cos\theta = 2\cos^{2}\left(\dfrac{\theta}{2}\right)$

Now,\left(\overset{\to}{a} + \overset{\to}{b}\right) \cdot \left(\overset{\to}{a} + \overset{\to}{b}\right) = {\left|\overset{\to}{a}\right|}^{2} + 2\overset{\to}{a} \cdot \overset{\to}{b} + {\left|\overset{\to}{b}\right|}^{2} ⇒ \left(\overset{\to}{a} + \overset{\to}{b}\right) \cdot \left(\overset{\to}{a} + \overset{\to}{b}\right) = {\left(1\right)}^{2} + 2 \times 1 \times 1 \times cos\theta + {\left(1\right)}^{2} ⇒ \left(\overset{\to}{a} + \overset{\to}{b}\right) \cdot \left(\overset{\to}{a} + \overset{\to}{b}\right) = 2\left(1 + cos\theta\right) ⇒ {\left|\overset{\to}{a} + \overset{\to}{b}\right|}^{2} = 4\cos^{2}\left(\dfrac{\theta}{2}\right) ⇒ \left|\overset{\to}{a} + \overset{\to}{b}\right| = 2\cos \left(\dfrac{\theta}{2}\right) . . . ( 1 )$</p><p>Similarly,</p><p>\left(\overset{\to}{a} - \overset{\to}{b}\right) \cdot \left(\overset{\to}{a} - \overset{\to}{b}\right) = {\left|\overset{\to}{a}\right|}^{2} - 2\overset{\to}{a} \cdot \overset{\to}{b} + {\left|\overset{\to}{b}\right|}^{2} ⇒ \left(\overset{\to}{a} - \overset{\to}{b}\right) \cdot \left(\overset{\to}{a} - \overset{\to}{b}\right) = {\left(1\right)}^{2} - 2 \times 1 \times 1 \times cos\theta + {\left(1\right)}^{2} ⇒ \left(\overset{\to}{a} - \overset{\to}{b}\right) \cdot \left(\overset{\to}{a} - \overset{\to}{b}\right) = 2\left(1 - cos\theta\right) ⇒ {\left|\overset{\to}{a} - \overset{\to}{b}\right|}^{2} = 4\sin^{2}\left(\dfrac{\theta}{2}\right) ⇒ \left|\overset{\to}{a} - \overset{\to}{b}\right| = 2\sin \left(\dfrac{\theta}{2}\right) . . . ( 2 )$

Therefore,

\left|\overset{^}{A} - \overset{^}{B}\right| = \left|\overset{^}{A} + \overset{^}{B}\right|\dfrac{tan\theta}{2}