JEE Main 2025 Jan 24 Shift 2, Mathematics Q3: Inverse Trigonometric Functions

Simplify each cotangent argument by combining over a common denominator. For the first term:

$\beta +\frac{1+{\beta }^2}{\alpha -\beta }=\frac{\beta (\alpha -\beta )+1+{\beta }^2}{\alpha -\beta }=\frac{\alpha \beta +1}{\alpha -\beta }$

Now use ${\cot }^{-1}\left(\frac{1+\alpha \beta }{\alpha -\beta }\right)={\tan }^{-1}\left(\frac{\alpha -\beta }{1+\alpha \beta }\right)={\tan }^{-1}\alpha -{\tan }^{-1}\beta$ (valid because $\alpha >\beta >0$, so the argument is positive).

Apply the same simplification to all three terms:

Term 1: ${\tan }^{-1}\alpha -{\tan }^{-1}\beta$

Term 2: ${\tan }^{-1}\beta -{\tan }^{-1}\gamma$

Term 3: ${\cot }^{-1}\left(\frac{\alpha \gamma +1}{\gamma -\alpha }\right)$. Here $\gamma <\alpha$ makes the argument negative. Using ${\cot }^{-1}(-x)=\pi -{\cot }^{-1}x$ for $x>0$:

$=\pi +{\tan }^{-1}\gamma -{\tan }^{-1}\alpha$

Adding all three terms causes every inverse tangent to cancel:

$({\tan }^{-1}\alpha -{\tan }^{-1}\beta )+({\tan }^{-1}\beta -{\tan }^{-1}\gamma )+(\pi +{\tan }^{-1}\gamma -{\tan }^{-1}\alpha )=\pi$

Correct option: (4)