If θ is the angle between unit vectors A→,B→ then (1−A→.B→)(1+A→.B→) is equal to

1−A→.B→1+A→.B→=1−ABcosθ1+ABcosθ=1−cosθ1+cosθ =2sin2(θ/2)2cos2(θ/2) =tan2(θ/2)

If θ is the angle between unit vectors A→,B→ then (1−A→.B→)(1+A→.B→) is equal to

1−A→.B→1+A→.B→=1−ABcosθ1+ABcosθ=1−cosθ1+cosθ =2sin2(θ/2)2cos2(θ/2) =tan2(θ/2)

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If $θ$ is the angle between unit vectors $\vec{A}, \vec{B}$ then $\frac{(1 - \vec{A} . \vec{B})}{(1 + \vec{A} . \vec{B})}$ is equal to

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$\cos^{2} (θ / 2)$

$\tan^{2} (θ / 2)$

$\cot^{2} (θ / 2)$

$\sin^{2} (θ / 2)$

answer is A.

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$\frac{1 - \vec{A} . \vec{B}}{1 + \vec{A} . \vec{B}} = \frac{1 - A B \cos θ}{1 + A B \cos θ} = \frac{1 - \cos θ}{1 + \cos θ} = \frac{2 \sin^{2} (θ / 2)}{2 \cos^{2} (θ / 2)} = \tan^{2} (θ / 2)$

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If θ is the angle between unit vectors A→,B→ then (1−A→.B→)(1+A→.B→) is equal to

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If $θ$ is the angle between unit vectors $\vec{A}, \vec{B}$ then $\frac{(1 - \vec{A} . \vec{B})}{(1 + \vec{A} . \vec{B})}$ is equal to