Question

If $4\cos \alpha =3\cos (\alpha -2\beta ),$ then $\tan (\alpha -\beta ):\cot \beta$ is

• A. $7:1$
• B. $1:7$
• C. $3:4$
• D. $4:3$

Answer 1

Answer: (B)

$1:7$

The given equation is
$4\cos (\alpha -\beta +\beta )=3\cos (\alpha -\beta -\beta )$
put $\alpha -\beta =\gamma$
$4\cos (\gamma +\beta )=3\cos (\gamma -\beta )$
$\Rightarrow \frac{\cos (\gamma -\beta )}{\cos (\gamma +\beta )}=\frac{4}{3}$
$\Rightarrow \frac{\cos (\gamma -\beta )+\cos (\gamma +\beta )}{\cos (\gamma -\beta )-\cos (\gamma +\beta )}=\frac{7}{1}$
$\Rightarrow \frac{2\cos \gamma \cos \beta }{2\sin \gamma \sin \beta }=\frac{7}{1}$
$\Rightarrow \tan \gamma \tan \beta =\frac{1}{7}$
$\Rightarrow \tan (\alpha -\beta ):\cot \beta =1:7$