2 =3 2 -1 3- 2 , then = - Vidyadip

MCQ

$\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$, then $\tan \alpha=$

✓
$\sqrt{2} \tan \beta$
B
$\frac{\tan \beta}{\sqrt{2}}$
C
$\frac{\tan ^2 \beta}{\sqrt{2}}$
D
$\tan \beta$

✓

Answer

Correct option: A.

$\sqrt{2} \tan \beta$

(A)
$\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$
By componendo - dividendo, we get
$\frac{\cos 2 \alpha+1}{\cos 2 \alpha-1}=\frac{3 \cos 2 \beta-1+3-\cos 2 \beta}{3 \cos 2 \beta-1-(3-\cos 2 \beta)}$
$\Rightarrow \frac{2 \cos ^2 \alpha}{-2 \sin ^2 \alpha}=\frac{2+2 \cos 2 \beta}{4 \cos 2 \beta-4}$
$\Rightarrow \frac{-\cos ^2 \alpha}{\sin ^2 \alpha}=\frac{1+\cos 2 \beta}{2(\cos 2 \beta-1)}=\frac{2 \cos ^2 \beta}{-4 \sin ^2 \beta}$
$\Rightarrow \frac{\sin ^2 \alpha}{\cos ^2 \alpha}=\frac{2 \sin ^2 \beta}{\cos ^2 \beta} \Rightarrow \tan ^2 \alpha=2 \tan ^2 \beta$
$\Rightarrow \tan \alpha=\sqrt{2} \tan \beta$

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