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

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tan² ( θ / 2)

sin² (θ / 2)

cot² (θ/2)

cos² (θ/2)

answer is A.

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Detailed Solution

$\frac{{1 - \vec A.\vec B}}{{1 + \vec A.\vec B}} = \frac{{1 - AB\cos \theta }}{{1 + AB\cos \theta }}\;\;(A = 1,\;B = 1)$

$\frac{{1 - \cos \theta }}{{1 + \cos \theta }} = \frac{{2{{\sin }^2}\theta /2}}{{2{{\cos }^2}\theta /2}} = {\tan ^2}\theta /2$

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