Let $\alpha$ and $\beta$ be nonzero real numbers such that $2(\cos \beta-\cos \alpha)+\cos \alpha \cos \beta=1$. Then which of the following is/are true?

(A) $\tan \left(\frac{\alpha}{2}\right)+\sqrt{3} \tan \left(\frac{\beta}{2}\right)=0$

(B) $\sqrt{3} \tan \left(\frac{\alpha}{2}\right)+\tan \left(\frac{\beta}{2}\right)=0$

(C) $\tan \left(\frac{\alpha}{2}\right)-\sqrt{3} \tan \left(\frac{\beta}{2}\right)=0$

(D) $\sqrt{3} \tan \left(\frac{\alpha}{2}\right)-\tan \left(\frac{\beta}{2}\right)=0$

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Trigonometric Ratios and Identities

Let

α

and

β

be nonzero real numbers such that

2 (cos β - cos α) + cos α cos β = 1

. Then which of the following is/are true?

From the given equation:

2 (cos β - cos α) + cos α cos β = 1

cos β (2 + cos α) = 1 + 2 cos α

\frac{cos β}{1} = \frac{1 + 2 cos α}{2 + cos α}

Using componendo-dividendo:

\frac{cos β + 1}{cos β - 1} = \frac{3 ( cos α + 1 )}{cos α - 1}

\frac{2 cos ^{2} \frac{β}{2}}{- 2 sin ^{2} \frac{β}{2}} = \frac{3 \cdot 2 cos ^{2} \frac{α}{2}}{- 2 sin ^{2} \frac{α}{2}}

- cot^{2} \frac{β}{2} = - 3 cot^{2} \frac{α}{2}

tan^{2} \frac{α}{2} = 3 tan^{2} \frac{β}{2}

tan \frac{α}{2} = \pm 3 tan \frac{β}{2}

This gives options (A) and (C). However, the question was awarded bonus marks to all candidates.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options