ITF – Summation Series – JEE Main 24 January 2025 Shift 2

Solution:

Let the given expression be $E$. We simplify the argument of each inverse cotangent term.

For the first term:
$$ \beta + \frac{1+\beta^2}{\alpha-\beta} = \frac{\beta(\alpha-\beta) + 1 + \beta^2}{\alpha-\beta} = \frac{\alpha\beta – \beta^2 + 1 + \beta^2}{\alpha-\beta} = \frac{1+\alpha\beta}{\alpha-\beta} $$
Since $\alpha > \beta > 0$, the argument is positive.
$$ \cot^{-1}\left( \frac{1+\alpha\beta}{\alpha-\beta} \right) = \tan^{-1}\left( \frac{\alpha-\beta}{1+\alpha\beta} \right) = \tan^{-1}\alpha – \tan^{-1}\beta $$

Similarly, for the second term:
$$ \gamma + \frac{1+\gamma^2}{\beta-\gamma} = \frac{1+\beta\gamma}{\beta-\gamma} $$
Since $\beta > \gamma > 0$, the argument is positive.
$$ \cot^{-1}\left( \frac{1+\beta\gamma}{\beta-\gamma} \right) = \tan^{-1}\left( \frac{\beta-\gamma}{1+\beta\gamma} \right) = \tan^{-1}\beta – \tan^{-1}\gamma $$

For the third term:
$$ \alpha + \frac{1+\alpha^2}{\gamma-\alpha} = \frac{1+\alpha\gamma}{\gamma-\alpha} $$
Here, $\gamma < \alpha$, so the denominator $(\gamma-\alpha)$ is negative. The argument of $\cot^{-1}$ is negative. Recall that for $x < 0$, $\cot^{-1}(x) = \pi + \tan^{-1}(1/x)$. $$ \cot^{-1}\left( \frac{1+\alpha\gamma}{\gamma-\alpha} \right) = \pi + \tan^{-1}\left( \frac{\gamma-\alpha}{1+\alpha\gamma} \right) = \pi + (\tan^{-1}\gamma - \tan^{-1}\alpha) $$
Adding all three simplified terms:
$$ E = (\tan^{-1}\alpha – \tan^{-1}\beta) + (\tan^{-1}\beta – \tan^{-1}\gamma) + (\pi + \tan^{-1}\gamma – \tan^{-1}\alpha) $$

Canceling the terms:
$$ E = \pi $$

Ans. (4)