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Vectors $\vec{A}\And \vec{B}$ include an angle $\theta$ between them. If $\vec{A}+\vec{B}$ and $\vec{A}-\vec{B}$ respectively subtend angles $\alpha \And \beta$ with , $\vec{A}$ then $(\tan \alpha +\tan \beta )$ is
Option: 1
$\frac{AB\sin \theta }{{{A}^{2}}+{{B}^{2}}{{\cos }^{2}}\theta }$

Option: 2
$\frac{2AB\sin \theta }{{{A}^{2}}-{{B}^{2}}{{\cos }^{2}}\theta }$

Option: 3
$\frac{{{A}^{2}}{{\sin }^{2}}\theta }{{{A}^{2}}+{{B}^{2}}{{\cos }^{2}}\theta }$

Option: 4
$\frac{{{B}^{2}}{{\sin }^{2}}\theta }{{{A}^{2}}-{{B}^{2}}{{\cos }^{2}}\theta }$

Answers (1)

best_answer

$\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }\$

Where $\alpha$ is the angle made by the vector $\vec{A}+\vec{B}$ with $\vec{A}$

Similarly, $\tan \beta =\frac{B\sin \theta }{A-B\cos \theta }\$

Where $\beta$ is the angle made by the vector $\vec{A}-\vec{B}$ with $\vec{A}$

The angle between $\vec{A}$ and $-\vec{B}$ is ${{180}^{\circ }}-\theta$

$\tan \alpha +\tan \beta =\frac{B\sin \theta }{A+B\cos \theta }+\frac{B\sin \theta }{A-B\cos \theta } \\\\ \tan \alpha +\tan \beta =\frac{B\sin \theta (A-B\cos \theta )+B\sin \theta (A+B\cos \theta )}{(A-B\cos \theta )(A+B\cos \theta )} \\\\ \tan \alpha +\tan \beta =\frac{AB\sin \theta -{{B}^{2}}\sin \theta \cos \theta )+AB\sin \theta +{{B}^{2}}\sin \theta \cos \theta )}{(A-B\cos \theta )(A+B\cos \theta )} \\\\ \tan \alpha +\tan \beta =\frac{2AB\sin \theta \cos \theta )}{({{A}^{2}}-{{B}^{2}}{{\cos }^{2}}\theta )} \\$

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Rakesh

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