Animated Solution for Mathematics - Trigonometry: In a triangle $ABC$, let $AB=\sqrt{23},BC=3$ and $CA=4$. Then the value of $\frac{\cot A+\cot C}{\cot B}$ is_____.

Imagine you are standing in the middle of a triangle $ABC$. You are given the lengths of its three sides: $a=3$, $b=4$, and $c=\sqrt{23}$.

At first glance, you might feel the urge to reach for your calculator to find the angles. But stop! In the world of JEE Advanced, the most elegant path is rarely the one that involves brute-force calculation.

Instead, we are going to look for the hidden symmetry in the problem. We are tasked with evaluating the expression $\frac{\cot A+\cot C}{\cot B}$. This is a classic setup where trigonometry meets algebra, and our goal is to let the geometry do the heavy lifting for us.

Connecting Trigonometry to Algebra

To solve this, we need a bridge. We need a way to express the trigonometric ratios $\cot A$, $\cot B$, and $\cot C$ in terms of the side lengths $a$, $b$, and $c$.

This is where the powerful identity involving the area of the triangle, denoted by $\Delta$, comes into play:

Why is this formula so beautiful? Because it connects the angle directly to the squares of the sides. Notice the pattern: for $\cot A$, we subtract the square of the opposite side $a^2$ and add the squares of the adjacent sides $b^2$ and $c^2$.

This pattern holds for all three angles. It is a rhythmic, predictable structure that simplifies our lives immensely.

Watching the Area Vanish

Now, let us perform the substitution for all three terms:

When we plug these into our target expression, $\frac{\cot A+\cot C}{\cot B}$, something magical happens. The term $\frac{1}{4\Delta }$ appears in every single numerator and denominator.

We can factor it out and watch it vanish entirely! We are left with a purely algebraic expression:

A Moment of Clarity

Look at the numerator now. It is a beautiful display of cancellation. We have $-a^2$ and $+a^2$, which sum to zero, and $+c^2$ and $-c^2$, which also sum to zero.

What remains? We are left with $b^2+b^2$, which is simply $2b^2$. The denominator remains $a^2+c^2-b^2$.

Our expression has collapsed from a complex trigonometric ratio into the simple, elegant form:

Now, we finally bring in our given values: $a=3$, $b=4$, and $c=\sqrt{23}$. Squaring these, we get $a^2=9$, $b^2=16$, and $c^2=23$.

Substituting these into our simplified expression, the numerator becomes $2(16)=32$. The denominator becomes $9+23-16$.

Calculating the denominator, $9+23=32$, and $32-16=16$. Finally, we have:

We have arrived at the solution not by grinding through angles, but by understanding the underlying structure of the triangle. The final answer is 2.