Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of ${\cot }^{-1}\left(\frac{\sqrt{1+{\tan }^2(2)}-1}{\tan (2)}\right)-{\cot }^{-1}\left(\frac{\sqrt{1+{\tan }^2(\frac{1}{2})}+1}{\tan (\frac{1}{2})}\right)$ is equal to

Welcome, future engineers. Today, we stand before a problem that looks like a tangled knot of inverse trigonometry. It is designed to test not just your memory of formulas, but your conceptual maturity.

We are asked to evaluate the expression:

At first glance, it is intimidating. But remember, in JEE Advanced, complexity is often a mask for a simple, elegant identity waiting to be revealed. Let us peel back the layers.

The first thing that should catch your eye is the structure $\sqrt{1+{\tan }^2\theta }$. You know the identity $1+{\tan }^2\theta ={\sec }^2\theta$. So, the expression becomes $\sqrt{{\sec }^2\theta }$.

Here is where the trap lies. Many students rush to write $\sec \theta$. But you must pause. The square root of a square is the absolute value: $\sqrt{{\sec }^2\theta }=|\sec \theta |$.

This is the pivot point of the entire problem. If you miss this, the signs will betray you later.

Now, we must determine the sign of $\sec \theta$ for our two specific angles: $2$ radians and $\frac{1}{2}$ radians.

For $\theta =2$, we know that $\frac{\pi }{2}\approx 1.57$ and $\pi \approx 3.14$. Since $1.57<2<3.14$, the angle $2$ radians resides in the second quadrant. In the second quadrant, cosine is negative, and therefore secant is negative. Thus, $|\sec (2)|=-\sec (2)$.

For $\theta =\frac{1}{2}$, we are clearly in the first quadrant, where all trigonometric ratios are positive. Thus, $|\sec (\frac{1}{2})|=\sec (\frac{1}{2})$.

With the modulus resolved, we substitute back. For the first term, we have:

Now, we invoke the half-angle identities: $1+\cos (2\alpha )=2{\cos }^2(\alpha )$ and $\sin (2\alpha )=2\sin (\alpha )\cos (\alpha )$. Setting $\alpha =1$, we get:

The first term is now ${\cot }^{-1}(-\cot (1))$. Using the property ${\cot }^{-1}(-x)=\pi -{\cot }^{-1}(x)$, this becomes $\pi -{\cot }^{-1}(\cot (1))=\pi -1$.

For the second term, we follow the same path:

Using the half-angle identities with $\alpha =\frac{1}{4}$, we get:

Thus, the second term is ${\cot }^{-1}(\cot (\frac{1}{4}))=\frac{1}{4}$.

We have arrived at the finish line. Our expression $E$ is simply $(\pi -1)-\frac{1}{4}$.

Combining these, we get:

It is a beautiful result, isn't it? The complexity vanished, leaving behind a clean, rational number subtracted from $\pi$. This is the essence of mathematics: finding the order within the chaos.