MHB Arrange t_1, t_2, t_3 and t_4 in decreasing order

[anemone]

Let $0<x<45^{\circ}$. Arrange

$$t_1=(\tan x)^{\tan x}$$, $$t_2=(\tan x)^{\cot x}$$, $$t_3=(\cot x)^{\tan x}$$, and $$t_4=(\cot x)^{\cot x}$$

in decreasing order.

 

[Fernando Revilla]

Spoiler

If $0<x<45º$,
$$0<\tan x <1\text{ and }0<\tan x<\cot x,\text{ so }t_2=(\tan x)^{\cot x}<(\tan x)^{\tan x}=t_1$$
$$1<\cot x,\text{ so }t_3=(\cot x)^{\tan x}<(\cot x)^{\cot x}=t_4.$$
$$0<\tan x<\cot x,\text{ so }t_1=(\tan x)^{\tan x}<(\cot x)^{\tan x}=t_3.$$
We conclude, $t_2<t_1<t_3<t_4.$

 

[anemone]

Spoiler

If $0<x<45º$,
$$0<\tan x <1\text{ and }0<\tan x<\cot x,\text{ so }t_2=(\tan x)^{\cot x}<(\tan x)^{\tan x}=t_1$$
$$1<\cot x,\text{ so }t_3=(\cot x)^{\tan x}<(\cot x)^{\cot x}=t_4.$$
$$0<\tan x<\cot x,\text{ so }t_1=(\tan x)^{\tan x}<(\cot x)^{\tan x}=t_3.$$
We conclude, $t_2<t_1<t_3<t_4.$

Thanks for participating, Fernando Revilla and you got it absolutely correct, of course! Well done!(Clapping):)