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Q.

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

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a

sin² ( $θ$ /2)

b

cot² ( $θ$ /2)

c

cos² ( $θ$ /2)

d

tan² ( $θ$ /2)

answer is A.

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Detailed Solution

$\frac{1 - \vec{A} . \vec{B}}{1 + \vec{A} . \vec{B}} = \frac{1 - ABcosθ}{1 + ABcosθ} (∵ A = 1, B = 1)$

$\Rightarrow \frac{1 - cosθ}{1 + cosθ} = \frac{2 \sin^{2} (θ / 2)}{2 \cos^{2} (θ / 2)} = \tan^{2} (θ / 2)$

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