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Given $θ \in (0, π / 4)$ and $t_{1} = (\tan θ)^{\tan θ}$ , $t_{2} = (\tan θ)^{\cot θ}, t_{3} = (\cot θ)^{\tan θ}$ and $t_{4} = (\cot θ)^{\cot θ}$ then

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$t_{3} > t_{1} > t_{2} > t_{4}$

$t_{1} > t_{2} > t_{3} > t_{4}$

$t_{4} > t_{3} > t_{1} > t_{2}$

$t_{2} > t_{3} > t_{1} > t_{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} : θ \in (0, π / 4) \\ 0 < \tan θ < 1 and \cot θ > 1 \Rightarrow \log \cot θ > 0 \end{array}$

Now, $t_{1} = (\tan θ)^{\tan θ} \Rightarrow \log t_{1} = \tan θ \log (\tan θ)$

$\begin{array}{l} \Rightarrow \log t_{1} = \tan θ \log \frac{1}{\cot θ} = \tan θ [\log 1 - \log (\cot θ)] \\ = - \tan θ \log (\cot θ) \end{array}$

Similarly $\log t_{2} = - \cot θ \log (\cot θ)$

$\log t_{3} = \tan θ \log (\cot θ), \log t_{4} = \cot θ \log (\cot θ)$

As $\cot θ > \tan θ,$ we have

$\begin{aligned} \log t_{4} > \log t_{3} > \log t_{1} > \log t_{1} \\ \Rightarrow t_{4} > t_{3} > t_{1} > t_{2} . \end{aligned}$

Alternately $θ \in (0, \frac{π}{4}) \Rightarrow 0 < \tan θ < 1 \Rightarrow \cot θ > 1$

Take $\tan θ = \frac{1}{2} \Rightarrow \cot θ = 2, and we get t_{4} > t_{3} > t_{1} > t_{2}$

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