The sum and difference of two perpendicular vectors of equal lengths are

Given   $\left|\overrightarrow{A}\right|=\left|\overrightarrow{B}\right|$  or $A=B$ 

Sum:     $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

$\Rightarrow$        $\left|\overrightarrow{R}\right|=\sqrt{A^2+B^2}=\sqrt{2}A$

$\tan {\alpha }_1=\frac{B\sin \theta }{A+B\cos \theta }=\frac{2\sin {\displaystyle \frac{\theta }{2}}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}=\tan \frac{\theta }{2}\Rightarrow {\alpha }_1=\frac{\theta }{2}$

Difference:        $\overrightarrow{S}=\overrightarrow{A}-\overrightarrow{B}$

$\Rightarrow$        $\left|\overrightarrow{S}\right|=\sqrt{A^2+B^2}=\sqrt{2}A$

${\alpha }_1={45}^0,{\alpha }_2={45}^0$

Hence $\overrightarrow{R}$  and $\overrightarrow{S}$  will be perpendicular and also of equal lengths