Let $\theta\in (0, \frac{\pi}{4})$ and $t_1 = (\tan\theta)^{\tan\theta}$, $t_2 = (\tan\theta)^{\cot\theta}$, $t_3=(\cot\theta)^{\tan\theta}$...

Question

Let $\theta \in (0, \frac{\pi}{4})$ and $t_1 = (\tan\theta)^{\tan\theta}$, $t_2 = (\tan\theta)^{\cot\theta}$, $t_3=(\cot\theta)^{\tan\theta}$ and $t_4=(\cot\theta)^{\cot\theta}$, then show that $t_4 > t_3 > t_1 > t_2$.

I don't know how to start this question. Please help. thank you:)

Answer 1

for $\theta\in(0,45^\circ)$

$\tan \theta\in (0,1)$

lets take $\displaystyle \tan \theta=\frac{1}{2}$

$\displaystyle t_{1}=\bigg(\tan \theta\bigg)^{\tan \theta}=\frac{1}{\sqrt{2}}\approx 0.7$

$\displaystyle t_{2}=\bigg(\tan \theta\bigg)^{\cot \theta}=\frac{1}{2^2}=0.25$

$\displaystyle t_{3}=\bigg(\cot \theta\bigg)^{\tan\theta}=\sqrt{2}$

$\displaystyle t_{4}=\bigg(\cot\theta\bigg)^{\cot \theta}=2^2=4$

$$t_{4}>t_{3}>t_{1}>t_{2}.$$

Answer 2

Hint:

In $0<\theta<\dfrac\pi4,$ $$\cot\theta-\tan\theta=2\cot2\theta>0$$

$\implies\cot\theta>\tan\theta$ which is $>0$

Now if $a>b>0, a^a-a^b=a^b(a^{a-b}-1)>0$

and similarly $\left(\dfrac ab\right)^a>1\implies a^a> b^a$ and so on

Answer 3

Probably the easiest method would be

AM>=GM

=>>> $(\tanθ + \cotθ)/2 \ge (\tanθ*\cotθ)^{1/2}$

=>>> $\tanθ+\cotθ \ge (\tanθ*1/\tanθ)^{1/2}$

=2