JEE Main & Advanced Physics Vectors Subtraction of vectors

Since, \[\overrightarrow{A}-\overrightarrow{B}=\overrightarrow{A}+(-\overrightarrow{B})\]  and  

\[|\overrightarrow{A}+\overrightarrow{B}|\,=\, \sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] \[\Rightarrow \]

\[|\overrightarrow{A}-\overrightarrow{B}|\,=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \,({{180}^{o}}-\theta )}\]               

Since, \[\cos \,(180-\theta )=-\cos \theta \]

\[\Rightarrow \] \[|\overrightarrow{A}-\overrightarrow{B}|\,=\,\sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos \theta }\]

\[\tan {{\alpha }_{1}}=\frac{B\sin \theta }{A+B\cos \theta }\]

and \[\tan {{\alpha }_{2}}=\frac{B\sin \,(180-\theta )}{A+B\cos \,(180-\theta )}\]

But \[\sin (180-\theta )=\sin \theta \] and \[\cos (180-\theta )=-\cos \theta \] \[\Rightarrow \]

\[\tan {{\alpha }_{2}}=\frac{B\sin \theta }{A-B\cos \theta }\]