1. Let the two unit vectors be $\widehat A$ and $\widehat B$ with angle $\theta$ between them.

   Since they are unit vectors, $|\widehat A|=|\widehat B|=1$ and $\widehat A\cdot \widehat B = \cos\theta.$

2. Find $|\widehat A+\widehat B|$.

   Using $|\vec x|^2=\vec x\cdot \vec x,$ we get $|\widehat A+\widehat B|^2=(\widehat A+\widehat B)\cdot(\widehat A+\widehat B).$

   Expanding, $|\widehat A+\widehat B|^2=|\widehat A|^2+|\widehat B|^2+2\widehat A\cdot \widehat B$ $=1+1+2\cos\theta$ $=2(1+\cos\theta).$

   Now use the identity $1+\cos\theta=2\cos^2\frac{\theta}{2}.$

   So, $|\widehat A+\widehat B|^2=4\cos^2\frac{\theta}{2}$ $\Rightarrow |\widehat A+\widehat B|=2\cos\frac{\theta}{2}.$

3. Find $|\widehat A-\widehat B|$.

   Similarly, $|\widehat A-\widehat B|^2=(\widehat A-\widehat B)\cdot(\widehat A-\widehat B)$ $=|\widehat A|^2+|\widehat B|^2-2\widehat A\cdot \widehat B$ $=1+1-2\cos\theta$ $=2(1-\cos\theta).$

   Using $1-\cos\theta=2\sin^2\frac{\theta}{2},$ we get $|\widehat A-\widehat B|^2=4\sin^2\frac{\theta}{2}$ $\Rightarrow |\widehat A-\widehat B|=2\sin\frac{\theta}{2}.$

4. Compare the two expressions.

   \frac{|\widehat A-\widehat B|}{|\widehat A+\widehat B|}=rac{2\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}=\tan\frac{\theta}{2}.

   Therefore, $|\widehat A-\widehat B|=|\widehat A+\widehat B|\tan\frac{\theta}{2}.$

5. Check options:

   • A: $|\widehat A + \widehat B| = |\widehat A - \widehat B|\tan \frac{\theta}{2}$ This is incorrect.
   • B: $|\widehat A - \widehat B| = |\widehat A + \widehat B|\tan \frac{\theta}{2}$ This is correct.
   • C and D involving $\cos\frac{\theta}{2}$ are incorrect.

Hence, the correct option is B.