Subtraction of vectors
Category : JEE Main & Advanced
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 \[\cos (180-\theta )=-\cos \theta \] \[\Rightarrow \]
\[\tan {{\alpha }_{2}}=\frac{B\sin \theta }{A-B\cos \theta }\]
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