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Question
If \[\hat{ a } \text{ and } \hat{b }\] are unit vectors inclined at an angle θ, prove that
\[\tan\frac{\theta}{2} = \frac{\left| \hat{a} -\hat{b} \right|}{\left| \hat{a} + \hat{b} \right|}\]
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Solution
\[\text{ Given that } \hat{a}\text{ and } \hat{b}\text{ are unit vectors }.\]
\[So,\left| \hat{a} \right|=1,\left| \hat{b} \right|=1\]
\[\text{ We have }\]
\[ \left| \hat{a} + \hat{b} \right|^2 = \left| \hat{a} \right|^2 + \left| \hat{b} \right|^2 + 2 \hat{a} .\hat{ b} \]
\[ = 1 + 1 + 2 \left| \hat{a} \right| \left| \hat{b} \right| \cos \theta\]
\[ = 2 + 2\cos \theta\]
\[ \Rightarrow \cos\theta = \frac{\left| \hat{a} + \hat{b} \right|^2 - 2}{2} . . . \left( 1 \right)\]
\[ \left| \hat{a} - b \right|^2 = \left| \hat{a} \right|^2 + \left| \hat{b} \right|^2 - 2 \hat{a} . \hat{b} \]
\[ = 1 + 1 - 2 \left| \hat{a} \right| \left| \hat{b} \right| \cos \theta\]
\[ = 2 - 2\cos \theta\]
\[ \Rightarrow \cos\theta = \frac{2 - \left| \hat{a} - \hat{b} \right|^2}{2} . . . \left( 2 \right)\]
\[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}}\]
\[ = \sqrt{\frac{1 - \frac{2 - \left| \hat{a} - \hat{b} \right|^2}{2}}{2}}[\text{ From } (2)]\]
\[ = \sqrt{\frac{2 + \left| \hat{a} - \hat{b} \right|^2 - 2}{4}}\]
\[ = \sqrt{\frac{\left| \hat{a} - \hat{b} \right|^2}{4}}\]
\[ = \frac{1}{2}\left| \hat{a} - \hat{b} \right|\]
\[\text{ Now },\]
\[\tan \frac{\theta}{2} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} = \frac{\frac{1}{2}\left| \hat{a} - \hat{b} \right|}{\frac{1}{2}\left| \hat{a} + \hat{b} \right|} = \frac{\left| \hat{a} - \hat{b} \right|}{\left| \hat{a} + b \right|}\]
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