Trigonometrical Equations 1 Question 19

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

(2017 Adv.)

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

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

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

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

Show Answer

Answer:

Correct Answer: 19. (b, c)

Solution:

  1. We have, $2(\cos \beta-\cos \alpha)+\cos \alpha \cos \beta=1$

$$ \begin{array}{lc} \text { or } 4(\cos \beta-\cos \alpha)+2 \cos \alpha \cos \beta=2 \\ \begin{array}{lc} \Rightarrow & 1-\cos \alpha+\cos \beta-\cos \alpha \cos \beta \\ =3+3 \cos \alpha-3 \cos \beta-3 \cos \alpha \cos \beta \\ \Rightarrow \quad & (1-\cos \alpha)(1+\cos \beta)=3(1+\cos \alpha)(1-\cos \beta) \\ \Rightarrow & \frac{(1-\cos \alpha)}{(1+\cos \alpha)}=\frac{3(1-\cos \beta)}{1+\cos \beta} \\ \Rightarrow & \tan ^{2} \frac{\alpha}{2}=3 \tan ^{2} \frac{\beta}{2} \\ \therefore & \tan \frac{\alpha}{2} \pm \sqrt{3} \tan \frac{\beta}{2}=0 \end{array} \end{array} $$



NCERT Chapter Video Solution

Dual Pane