If $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are two vectors of equal magnitude and $\displaystyle \alpha$ is the angle between them, then prove that $\displaystyle \frac{|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|}{|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|}=\cot \left(\frac{\alpha}{2}\right)$.

Consider \(\displaystyle \frac{|\vec{a}+\vec{b}|^{2}}{|\vec{a}-\vec{b}|^{2}}=\frac{|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a}||\vec{b}| \cos \alpha}{|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}||\vec{b}| \cos \alpha}\) \[\begin{aligned} & =\frac{2 m^{2}(1+\cos \alpha)}{2 m^{2}(1-\cos \alpha)} \quad \text { where }|\vec{a}|=|\vec{b}|=m \\ & =\frac{2 \cos ^{2} \dfrac{\alpha}{2}}{2 \sin ^{2} \dfrac{\alpha}{2}} \\ & =\cot ^{2}\left(\frac{\alpha}{2}\right) \end{aligned} \] \[\therefore \frac{|\vec{a}+\vec{b}|}{|\vec{a}-\vec{b}|}=\cot \left(\frac{\alpha}{2}\right) \]