Question

If tan $\alpha =3\tan \beta$, then the maximum value ${\tan }^2(\alpha -\beta )$ is

• A. $\frac{4}{3}$
• B. $\frac{1}{3}$
• C. $\frac{49}{12}$
• D. $\frac{25}{12}$

Answer 1

Answer: (B)

$\frac{1}{3}$

$\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }=\frac{2\tan \beta }{1+3{\tan }^2\beta }=\frac{2}{\cot \beta +3\tan \beta }$
${\tan }^2(\alpha -\beta )=\frac{4}{(\cot \beta +3\tan \beta {)}^2}=\frac{4}{(\cot \beta -3\tan \beta {)}^2+12}\le \frac{1}{3}$
Maximum value $=\frac{1}{3}$