Question

Vectors $\vec{A}$ and $\vec{B}$ include an angle $\theta$ between them. If $\left(\vec{A}+\vec{B}\right)$ and $\left(\vec{A}-\vec{B}\right)$ respectively subtend angles $\alpha$ and $\beta$ with $\vec{A}$, then $\left(tan\,\alpha+tan\,\beta\right)$ is

• A. $\frac{\left(AB\,sin\,\theta\right)}{\left(A^{2}+B^{2}\,cos^{2}\,\theta\right)}$
• B. $\frac{\left(2AB\,sin\,\theta\right)}{\left(A^{2}-B^{2}\,cos^{2}\,\theta\right)}$
• C. $\frac{\left(A^{2}\,sin^{2}\,\theta\right)}{\left(A^{2}+B^{2}\,cos^{2}\,\theta \right)}$
• D. $\frac{\left(B^{2}\,sin^{2}\,\theta\right)}{\left(A^{2}-B^{2}\,cos^{2}\,\theta \right)}$

Answer 1

Answer: (B)

$tan\,\alpha=\frac{B\,sin\,\theta}{A+B\,cos\,\theta}\,...\left(i\right)$
where $\alpha$ is the angle made by the vector $\left(\vec{A}+\vec{B}\right)$ with $\vec{A}$.
Similarly, $tan\,\beta=\frac{B\,sin\,\theta}{A-B\,cos\,\theta}\,...\left(ii\right)$
where $\beta$ is the angle made by the vector $\left(\vec{A}-\vec{B}\right)$ with $\vec{A}$.
Note that the angle between $\vec{A}$. and $\left(-\vec{B}\right)$ is $\left(180^{\circ}-\theta\right)$.
Adding (i) and (ii), we get
$tan\,\alpha+tan\,\beta=\frac{B\,sin\,\theta}{A+B\,cos\,\theta}+\frac{B\,sin\,\theta}{A-B\,cos\,\theta}$
$=\frac{AB\,sin\,\theta-B^{2}\,sin\,\theta\,cos\,\theta+AB\,sin\,\theta+B^{2}\,sin\,\theta\,cos\,\theta}{\left(A+B\,cos\,\theta\right)\left(A-B\,cos\,\theta\right)}$
$=\frac{2AB\,sin\,\theta}{\left(A^{2}-B^{2}\,cos^{2}\,\theta\right)}$