If tan⁡θ=ntan⁡ϕthen the maximum value of tan2⁡(θ−ϕ)is equal to

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If $\tan θ = ntan ϕ$ then the maximum value of $\tan^{2} (θ - ϕ)$ is equal to

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$\frac{(n - 1)^{2}}{2 n}$

$\frac{(n + 1)^{2}}{4 n}$

$\frac{(n + 1)^{2}}{2 n}$

$\frac{(n - 1)^{2}}{4 n}$

answer is B.

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Detailed Solution

$\begin{array}{l} \tan (θ - ϕ) = \frac{\tan θ - \tan ϕ}{1 + \tan θtan ϕ} = \frac{(n - 1) \tan ϕ}{1 + {ntan}^{2} ϕ} \\ \Rightarrow \tan (θ - ϕ) = \frac{n - 1}{\cot ϕ + ntan ϕ} \\ \Rightarrow \cot ϕ + ntan ϕ \geq 2 \sqrt{n} \\ \tan^{2} (θ - ϕ) = \frac{(n - 1)^{2}}{{(\cot ϕ + {ntan}^{2} ϕ)}^{2}} \\ \Rightarrow maximum value of \tan^{2} (θ - ϕ) = \frac{(n - 1)^{2}}{4 n} \end{array}$