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Let $θ \in (0, \frac{π}{2})$ and $t_{1} = {(\cos e c θ)}^{\sin θ}, t_{2} = {(\sin θ)}^{\cos e c θ}, t_{3} = {(cosec θ)}^{\cos e c θ}, t_{4} = {(\sin θ)}^{\sin θ}$ , then

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

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

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

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

answer is D.

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Detailed Solution

t_{1} = {(\cos e c θ)}^{\sin θ}

$\log t_{1} = \sin θ \log (cosec θ)$

$= \sin θ \log (\frac{1}{\sin θ})$

$= - \sin θ \log (\sin θ) \to (1)$

$t_{2} = {(\sin θ)}^{cosec θ}$

$\log t_{2} = cosec θ \log (\sin θ) \to (2)$

$t_{3} = {(cosec θ)}^{cosec θ}$

$\log t_{3} = - cosec θ \log (\sin θ) \to (3)$

$\log t_{4} = \sin θ \log (\sin θ) \to (4)$

$∴$ if $θ \in (0, \frac{π}{2})$ $0 < \sin θ < 1$

$∴ \log (\sin θ) < 0$

From $(1) & (2)$

$\log t_{1} > \log t_{2} \Rightarrow t_{1} > t_{2}$

From $(2) & (3)$ $t_{3} > t_{2}$

From $(3) & (4)$ $t_{3} > t_{4}$

From $(1) & (4)$ $t_{1} > t_{4}$

From $(1) & (3)$ $t_{3} > t_{1}$

$∴ t_{3} > t_{1} > t_{4} > t_{2}$

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