Let and be nonzero real numbers such that 2( - )+ =1. Then which … - Vidyadip

MCQ

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?

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

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

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

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

✓
$A,C$
B
$A,B$
C
$A,D$
D
$A,C,D$

✓

Answer

Correct option: A.

$A,C$

a
$\frac{\cos \beta(2+\cos \alpha)=1+2 \cos \alpha}{1}=\frac{1+2 \cos \alpha}{2+\cos \alpha}$

$\frac{\cos \beta+1}{\cos \beta-1}=\frac{3(\cos \alpha+1)}{\cos \alpha-1}$

$\frac{2 \cos ^2 \frac{\beta}{2}}{-2 \sin ^2 \frac{\beta}{2}}=\frac{3 \times 2 \cos ^2 \frac{\alpha}{2}}{-2 \sin ^2 \frac{\alpha}{2}}$

$\tan ^2 \frac{\alpha}{2}=3 \tan ^2 \frac{\beta}{2}$

$\Rightarrow \quad \tan \frac{\alpha}{2}= \pm \sqrt{3} \tan \frac{\beta}{2}$

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