If tan θ = 3tan ϕ , then the maximum value of ^ 2 (θ -ϕ) is (where,...

tan \theta = 3tan ϕ

, then the maximum value of

\tan^{2}

\left(\theta -ϕ\right)

is (where,

tan ϕ > 0)

Options

A $1$
B $\dfrac{1}{3}$
C $2$
D $4$

Correct answer

B. $\dfrac{1}{3}$

Step-by-step solution

$tan \theta = 3tan ϕ$
$\tan \left(\theta - ϕ\right) = \dfrac{tan \theta - tan ϕ}{1 + tan \theta tan ϕ}$
$= \dfrac{2tan ϕ}{1 + 3\tan^{2} ϕ} = \dfrac{2}{cot ϕ + 3tan ϕ}$
Using $AM \ge GM$

$\dfrac{cot ϕ + 3tan ϕ}{2} \ge \sqrt{3}$
$⇒ \dfrac{2}{cot ϕ + 3tan ϕ} \le \dfrac{1}{\sqrt{3}}$
$⇒ \tan \left(\theta - ϕ\right) \le \dfrac{1}{\sqrt{3}}$
$⇒ \tan^{2} \left(\theta - ϕ\right) \le \dfrac{1}{3}$

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