Trigonometrical Equations Ques 12
- 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) $\sqrt{3} \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}=0$
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Answer:
Correct Answer: 12.(b, c)
Solution:
Formula:
- We have, $2(\cos \beta-\cos \alpha)+\cos \alpha \cos \beta=1$
$$ \begin{array}{lc} \text { or } 4(\cos \beta-\cos \alpha)+2 \cos \alpha \cos \beta=2 \\ \begin{array}{lc} \Rightarrow & 1-\cos \alpha+\cos \beta-\cos \alpha \cos \beta \\ =3+3 \cos \alpha-3 \cos \beta-3 \cos \alpha \cos \beta \\ \Rightarrow \quad & (1-\cos \alpha)(1+\cos \beta)=3(1+\cos \alpha)(1-\cos \beta) \\ \Rightarrow & \frac{(1-\cos \alpha)}{(1+\cos \alpha)}=\frac{3(1-\cos \beta)}{1+\cos \beta} \\ \Rightarrow & \tan ^{2} \frac{\alpha}{2}=3 \tan ^{2} \frac{\beta}{2} \\ \therefore & \tan \frac{\alpha}{2} \pm \sqrt{3} \tan \frac{\beta}{2}=0 \end{array} \end{array} $$
BETA
