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Arrange t_1, t_2, t_3 and t_4 in decreasing order

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Let $0<x<45^{\circ}$. Arrange

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

in decreasing order.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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, $t_2<t_1<t_3<t_4.$
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
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):)