Question
Let $$\theta \in \left( {0,\frac{\pi }{4}} \right)$$ and $${t_1} = {\left( {\tan \theta } \right)^{\tan \theta }},{t_2} = {\left( {\tan \theta } \right)^{\cot \theta }},$$ $${t_3} = {\left( {\cot \theta } \right)^{\tan \theta }}\,{\text{and }}{t_4} = {\left( {\cot \theta } \right)^{\cot \theta }},$$ then
A.
$${t_1} > {t_2} > {t_3} > {t_4}$$
B.
$${t_4} > {t_3} > {t_1} > {t_2}$$
C.
$${t_3} > {t_1} > {t_2} > {t_4}$$
D.
$${t_2} > {t_3} > {t_1} > {t_4}$$
Answer :
$${t_4} > {t_3} > {t_1} > {t_2}$$
Solution :
$$\eqalign{
& \because \,\,\theta \in \left( {0,\frac{\pi }{4}} \right) \cr
& \Rightarrow \,\,\tan \theta < 1\,{\text{and }}\cot \theta > 1 \cr
& {\text{Let tan}}\theta = 1 - x\,{\text{and cot}}\theta = 1 + y \cr} $$
Where $$x, y > 0$$ and are very small, then
$$\eqalign{
& \therefore \,\,{t_1} = {\left( {1 - x} \right)^{1 - x}},{t_2} = {\left( {1 - x} \right)^{1 + y}}, \cr
& {t_3} = {\left( {1 + y} \right)^{1 - x}},{t_4} = {\left( {1 + y} \right)^{1 + y}} \cr} $$
$${\text{Clearly, }}{t_4} > {t_3}\,{\text{and }}{t_1} > {t_2}\,{\text{also, }}{t_3} > {t_1}$$ NOTE THIS STEP
$${\text{Thus}}\,{\text{ }}{t_4} > {t_3} > {t_1} > {t_2}.$$