Question

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:

• A. $\dfrac{ab\sin\theta}{a^2+b^2\cos^2\theta}$
• B. $\dfrac{2b\sin\theta}{a^2-b^2\cos^2\theta}$
• C. $\dfrac{a^2\sin^2\theta}{a^2+b^2\cos^2\theta}$
• D. $\dfrac{b^2\sin^2\theta}{a^2-b^2\cos^2\theta}$

Hint:

For angle problems involving vectors:
Use $\displaystyle \tan\phi=\frac{|\vec p\times\vec q|}{\vec p\cdot\vec q}$
Cross products give $\sin$, dot products give $\cos$
Symmetry in $(\vec a+\vec b)$ and $(\vec a-\vec b)$ often simplifies algebra

Answer 1

The Correct Option is B

Step 1: Recall the formula for angle between vectors. For vectors $\vec p$ and $\vec q$, the tangent of the angle $\phi$ between them is: \[ \tan\phi=\frac{|\vec p\times\vec q|}{\vec p\cdot\vec q} \]
Step 2: Find $\tan\alpha$. Angle $\alpha$ is between $\vec a$ and $(\vec a+\vec b)$: \[ \tan\alpha=\frac{|\vec a\times(\vec a+\vec b)|}{\vec a\cdot(\vec a+\vec b)} \] Since $\vec a\times\vec a=0$, \[ |\vec a\times(\vec a+\vec b)|=|\vec a\times\vec b|=ab\sin\theta \] Also, \[ \vec a\cdot(\vec a+\vec b)=a^2+ab\cos\theta \] Hence, \[ \tan\alpha=\frac{ab\sin\theta}{a^2+ab\cos\theta} \]
Step 3: Find $\tan\beta$. Angle $\beta$ is between $\vec a$ and $(\vec a-\vec b)$: \[ \tan\beta=\frac{|\vec a\times(\vec a-\vec b)|}{\vec a\cdot(\vec a-\vec b)} \] \[ |\vec a\times(\vec a-\vec b)|=ab\sin\theta \] \[ \vec a\cdot(\vec a-\vec b)=a^2-ab\cos\theta \] Thus, \[ \tan\beta=\frac{ab\sin\theta}{a^2-ab\cos\theta} \]
Step 4: Add $\tan\alpha+\tan\beta$. \[ \tan\alpha+\tan\beta = ab\sin\theta\left(\frac{1}{a^2+ab\cos\theta}+\frac{1}{a^2-ab\cos\theta}\right) \] \[ = ab\sin\theta\cdot\frac{2a^2}{a^4-a^2b^2\cos^2\theta} \] \[ = \frac{2b\sin\theta}{a^2-b^2\cos^2\theta} \] Final Answer: \[ \boxed{\tan\alpha+\tan\beta=\frac{2b\sin\theta}{a^2-b^2\cos^2\theta}} \]