If $~A~$ is between $~0^{\circ}~$ and $~45^{\circ}~$, solve the following

Question

If $~A~$ is between $~0^{\circ}~$ and $~45^{\circ}~$, $$T_1=\tan A^{\tan A}$$ $$T_2=\tan A^{\cot A}$$ $$T_3=\cot A^{\tan A}$$ $$T_4=\cot A^{\cot A}$$

Arrange them in accessing order

Answer 1

For $$x\in(0,45^\circ) \ \ , 0\lt \tan x \lt1\text{ and }1\lt \cot x\lt \infty \implies \tan x \lt \cot x $$

For a positive number $z$ and $x \in (0,45^\circ)$, $(\tan x)^z \lt (\cot x)^z$

$\bullet$ Now, for a number say $k \in (0,1)$ greater the positive power $h$ you apply, smaller the value of $k^h$ will be.

So, $$0 \lt(\tan x)^{\cot x} \lt (\tan x)^{\tan x} \lt 1 $$

$\bullet$ And, for a number say $k \in (1,\infty)$ greater the positive power $h$ you apply, greater the value of $k^h$ will be.

So, $$1 \lt(\cot x)^{\tan x} \lt (\cot x)^{\cot x} \lt \infty$$

Thus

$$(\tan x)^{\cot x} \lt (\tan x)^{\tan x}\lt(\cot x)^{\tan x} \lt (\cot x)^{\cot x}$$

Answer 2

If $0<A<45^\circ,0<\tan A<1,\cot A>1$

Take logarithm wrt $\cot A$

$$\log_{\cot A} T_1=-\tan A,\log_{\cot A} T_2=-\cot A,\log_{\cot A} T_3=\tan A,\log_{\cot A}T_4=\cot A$$

Use $\cot A>\tan A$

We can prove if $\log A>\log B\iff A>B$ for base $>1$

Answer 3

Write $\tan A=x,\cot A=y, 0<x<1;y>1$

$x^y-x^x=x^x(x^{y-x}-1)<0$ as $y-x>0$ and $0<x<1$

As $x<y, x^x<y^y$

$y^y-y^x=y^x(y^{y-x}-1)>0$ as $y>1,y-x>0$

$y^y-x^y=x^y\left(\left(\dfrac yx\right)^y-1\right)>0$ as $\dfrac yx>1,y>1$