Animated Solution for Mathematics - Trigonometry: Let $\theta \in (0,\pi /4)$ 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

Imagine you are standing on the unit circle, looking at the angle $\theta$ as it sweeps from $0$ to $\frac{\pi }{4}$. This is the playground for our problem.

We are given four terms: $t_1=(\tan \theta {)}^{\tan \theta }$, $t_2=(\tan \theta {)}^{\cot \theta }$, $t_3=(\cot \theta {)}^{\tan \theta }$, and $t_4=(\cot \theta {)}^{\cot \theta }$.

At first glance, these look like a chaotic mess of trigonometric functions. The secret to mastering JEE problems is not brute force; it is finding the right perspective.

Let's simplify our life. We know that for $\theta \in (0,\frac{\pi }{4})$, $\tan \theta$ is a value strictly between $0$ and $1$.

Let's define $x=\tan \theta$, where $0<x<1$. Consequently, $\cot \theta =\frac{1}{\tan \theta }=\frac{1}{x}$.

Since $x$ is a fraction less than $1$, its reciprocal $\frac{1}{x}$ is definitely greater than $1$. Now, our terms look much cleaner:

This is the algebraic soul of the problem.

We have two exponents in play: $x$ and $\frac{1}{x}$. Since $x$ is between $0$ and $1$, it is a mathematical certainty that $x<\frac{1}{x}$.

This simple inequality is the lever that will move the world. Now, we must analyze the behavior of the exponential function $f(y)=a^y$.

The behavior depends entirely on the base $a$. If $a>1$, the function is strictly increasing. If $0<a<1$, the function is strictly decreasing.

Let's look at $t_3=(\frac{1}{x}{)}^x$ and $t_4=(\frac{1}{x}{)}^{1/x}$. Here, the base is $\frac{1}{x}$, which is greater than $1$.

Because the base is greater than $1$, the function is increasing. Since the exponent of $t_4$ (which is $\frac{1}{x}$) is greater than the exponent of $t_3$ (which is $x$), it follows that $t_4>t_3$.

Furthermore, since the base is greater than $1$ and the exponent is positive, both $t_3$ and $t_4$ are greater than $1$. So, we have:

Now, let's look at $t_1=x^x$ and $t_2=x^{1/x}$. Here, the base is $x$, which is less than $1$.

This is where the trap lies! When the base is between $0$ and $1$, the function is strictly decreasing. This means a larger exponent results in a smaller value.

Since the exponent of $t_2$ (which is $\frac{1}{x}$) is greater than the exponent of $t_1$ (which is $x$), the inequality reverses: $t_1>t_2$. Also, because the base is less than $1$ and the exponent is positive, both $t_1$ and $t_2$ are less than $1$. Thus, we have:

We have two beautiful chains of inequalities. We know $t_4>t_3>1$ and $1>t_1>t_2$.

By using the number $1$ as our bridge, we can connect these two chains into one magnificent sequence:

This leads us directly to the conclusion that $t_4>t_3>t_1>t_2$. You have just navigated the treacherous waters of exponential inequalities.