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

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 $$t_4 > t_3 > t_1 > t_2$$.

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

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

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

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

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

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

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

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

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