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

1−A¯.B¯1+A¯.B¯=1−ABcosθ1+ABcosθ=1−A(A)cosθ1+AAcosθ=1−(1)(1)cosθ1+(1)(1)cosθ=1−cosθ1+cosθ=2sin2θ/22cos2θ/2=tan2θ/2

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

1−A¯.B¯1+A¯.B¯=1−ABcosθ1+ABcosθ=1−A(A)cosθ1+AAcosθ=1−(1)(1)cosθ1+(1)(1)cosθ=1−cosθ1+cosθ=2sin2θ/22cos2θ/2=tan2θ/2

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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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$\tan^{2} (θ /2)$

$\sin^{2} (θ /2)$

$\cot^{2} (θ /2)$

$\cos^{2} (θ /2)$

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$\frac{1 - \bar{A} . \bar{B}}{1 + \bar{A} . \bar{B}} = \frac{1 - A B \cos θ}{1 + A B \cos θ}$

$= \frac{1 - A (A) \cos θ}{1 + A A \cos θ}$

$= \frac{1 - (1) (1) \cos θ}{1 + (1) (1) \cos θ}$

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

$= \tan^{2} θ /2$

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

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