Question

If $a$ and $b$ unit vectors inclined at an angle $\theta$, then prove that :

$\tan \frac{\theta }{2}=\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$

Answer 1

Using the formula:
$|\overrightarrow{A}\pm \overrightarrow{B}|=\sqrt{|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2\pm 2|\overrightarrow{A}||\overrightarrow{B}|\cos \theta }$
$|\hat{a}+\hat{b}|=\sqrt{1+1+2\cos \theta }=\sqrt{2(1+\cos \theta )}=\sqrt{4{\cos }^2\frac{\theta }{2}}$
$|\hat{a}-\hat{b}|=\sqrt{1+1-2\cos \theta }=\sqrt{2(1-\cos \theta )}=\sqrt{4{\sin }^2\frac{\theta }{2}}$
$\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}=\frac{\sqrt{4{\sin }^2\frac{\theta }{2}}}{\sqrt{4{\cos }^2\frac{\theta }{2}}}=\tan \frac{\theta }{2}$