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- Feb 14, 2012

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

in decreasing order.

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- Thread starter
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- #1

- Feb 14, 2012

- 3,917

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

in decreasing order.

- Jan 29, 2012

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$$0<\tan x <1\text{ and }0<\tan x<\cot x,\text{ so }t_2=(\tan x)^{\cot x}<(\tan x)^{\tan x}=t_1$$

$$1<\cot x,\text{ so }t_3=(\cot x)^{\tan x}<(\cot x)^{\cot x}=t_4.$$

$$0<\tan x<\cot x,\text{ so }t_1=(\tan x)^{\tan x}<(\cot x)^{\tan x}=t_3.$$

We conclude, $t_2<t_1<t_3<t_4.$

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- Feb 14, 2012

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Thanks for participating,

$$0<\tan x <1\text{ and }0<\tan x<\cot x,\text{ so }t_2=(\tan x)^{\cot x}<(\tan x)^{\tan x}=t_1$$

$$1<\cot x,\text{ so }t_3=(\cot x)^{\tan x}<(\cot x)^{\cot x}=t_4.$$

$$0<\tan x<\cot x,\text{ so }t_1=(\tan x)^{\tan x}<(\cot x)^{\tan x}=t_3.$$

We conclude, $t_2<t_1<t_3<t_4.$