Question

If $\tan \theta=3 \tan \phi$, then the maximum value of $\tan ^{2}(\theta-\phi)$ is (where, $\tan \phi>0)$

• A. $1$
• B. $\frac{1}{3}$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$\tan \theta=3 \tan \phi$
$\tan (\theta-\phi)=\frac{\tan \theta-\tan \phi}{1+\tan \theta \tan \phi}$
$=\frac{2 \tan \phi}{1+3 \tan ^{2} \phi}=\frac{2}{\cot \phi+3 \tan \phi}$
Using AM $\geq$ GM
$\frac{\cot \phi+3 \tan \phi}{2} \geq \sqrt{3}$
$\Rightarrow \quad \frac{2}{\cot \phi+3 \tan \phi} \leq \frac{1}{\sqrt{3}}$
$\Rightarrow \tan (\theta-\phi) \leq \frac{1}{\sqrt{3}}$
$\Rightarrow \tan ^{2}(\theta-\phi) \leq \frac{1}{3}$