Animated Solution for Mathematics - Inverse Trigonometric Functions: If $\alpha >\beta >\gamma >0,$ then the expression ${\cot }^{-1}\{\beta +\frac{(1+{\beta }^2)}{(\alpha -\beta )}\}+{\cot }^{-1}\{\gamma +\frac{(1+{\gamma }^2)}{(\beta -\gamma )}\}+{\cot }^{-1}\{\alpha +\frac{(1+{\alpha }^2)}{(\gamma -\alpha )}\}$ is equal to:

Welcome, aspiring engineer. Today, we are going to dismantle a problem that, at first glance, looks like a tangled web of inverse trigonometric functions. It is designed to intimidate, but beneath the surface, it is a masterpiece of symmetry.

We are given the expression:

This is subject to the strict constraint $\alpha >\beta >\gamma >0$. This constraint is our North Star; without it, we would be lost in a sea of undefined values.

A Pattern Emerges

Let us isolate the first term, $T_1={\cot }^{-1}\left(\beta +\frac{1+{\beta }^2}{\alpha -\beta }\right)$. To simplify this, we take the common denominator inside the bracket:

The $-{\beta }^2$ and $+{\beta }^2$ terms vanish, leaving us with ${\cot }^{-1}\left(\frac{1+\alpha \beta }{\alpha -\beta }\right)$. Since $\alpha >\beta >0$, the argument is positive, allowing us to use the identity ${\cot }^{-1}x={\tan }^{-1}(1/x)$.

This transforms $T_1$ into:

The second term, $T_2$, follows the exact same logic. By symmetry, we find:

The Third Term

Now, we arrive at the third term, $T_3={\cot }^{-1}\left(\alpha +\frac{1+{\alpha }^2}{\gamma -\alpha }\right)$. Applying the same algebraic steps:

Here is where the JEE examiner is watching you. Look at the denominator: $\gamma -\alpha$. Because $\alpha >\gamma$, this denominator is negative, meaning the argument of the ${\cot }^{-1}$ function is negative.

We must use the property ${\cot }^{-1}x=\pi +{\tan }^{-1}(1/x)$ for $x<0$. Therefore:

We now combine our three pieces: $T_1={\tan }^{-1}\alpha -{\tan }^{-1}\beta$, $T_2={\tan }^{-1}\beta -{\tan }^{-1}\gamma$, and $T_3=\pi +{\tan }^{-1}\gamma -{\tan }^{-1}\alpha$. Summing them up:

Watch the magic: ${\tan }^{-1}\alpha$ cancels with $-{\tan }^{-1}\alpha$, $-{\tan }^{-1}\beta$ cancels with ${\tan }^{-1}\beta$, and $-{\tan }^{-1}\gamma$ cancels with ${\tan }^{-1}\gamma$. We are left with exactly:

You have successfully navigated the trap. The lesson here is simple but profound: never blindly apply identities. Always respect the domain and the sign of the argument.