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

Given that $\overrightarrow{\mathrm{a}}$and $\overrightarrow{\mathrm{b}}$ are having equal magnitude,
So, let $\left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{b}}\right|=\mathrm{x}$
Now, 
$\begin{array}{l}Now \frac{|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}{|}^2}{|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}{|}^2}\\ =\frac{|\overrightarrow{\mathrm{a}}{|}^2+|\overrightarrow{\mathrm{b}}{|}^2+2\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\cos \mathrm{\alpha }}{|\overrightarrow{\mathrm{a}}{|}^2+|\overrightarrow{\mathrm{b}}{|}^2-2\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\cos \mathrm{\alpha }}\\ \left.=\frac{{\mathrm{x}}^2+{\mathrm{x}}^2+2{\mathrm{x}}^2\cos \mathrm{\alpha }}{{\mathrm{x}}^2+{\mathrm{x}}^2-2{\mathrm{x}}^2\cos \mathrm{\alpha }}\text{ (Because, }\left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{b}}\right|=\mathrm{x}\right)\\ =\frac{2{\mathrm{x}}^2+2{\mathrm{x}}^2\cos \mathrm{\alpha }}{2{\mathrm{x}}^2-2{\mathrm{x}}^2\cos \mathrm{\alpha }}\\ =\frac{2{\mathrm{x}}^2(1+\cos \mathrm{\alpha })}{2{\mathrm{x}}^2(1-\cos \mathrm{\alpha })}\\ =\frac{2{\cos }^2\frac{\mathrm{\alpha }}{2}}{2{\sin }^2\frac{\mathrm{\alpha }}{2}}\left(\because 1+\cos \mathrm{\alpha }=2{\cos }^2\frac{\mathrm{\alpha }}{2}\text{ and }1-\cos \mathrm{\alpha }=2{\sin }^2\mathrm{\alpha }\right)\\ ={\cot }^2\frac{\mathrm{\alpha }}{2}\end{array}$
$\therefore \frac{|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|}{|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|}=\cot \frac{\mathrm{\alpha }}{2}$

That have proven.