If $\alpha>\beta>\gamma>0$, then the expression $\cot ^{-1}\left\{\beta+\frac{\left(1+\beta^{2}\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^{2}\right)}{(\beta-\gamma)}\right\}$ $+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^{2}\right)}{(\gamma-\alpha)}\right\}$ is equal to:

(A) $\frac{\pi}{2}-(\alpha+\beta+\gamma)$

Inverse Trigonometric Functions — JEE Main · 2025 · 24 Jan 2025 · Shift 2 Question with Solution

α > β > γ > 0

, then the expression

cot^{- 1} {β + \frac{( 1 + β ^{2} )}{( α - β )}} + cot^{- 1} {γ + \frac{( 1 + γ ^{2} )}{( β - γ )}}

+ cot^{- 1} {α + \frac{( 1 + α ^{2} )}{( γ - α )}}

is equal to:

Answer: (D)

Simplify each cotangent term by writing it as a tangent of a difference. Take the first term:

cot^{- 1} {β + \frac{1 + β ^{2}}{α - β}} = cot^{- 1} (\frac{α β + 1}{α - β}) = tan^{- 1} (\frac{α - β}{1 + α β}) = tan^{- 1} α - tan^{- 1} β,

using the identity

tan^{- 1} α - tan^{- 1} β = tan^{- 1} (\frac{α - β}{1 + α β})

, valid here since

α > β

.

Similarly, since

β > γ

cot^{- 1} {γ + \frac{1 + γ ^{2}}{β - γ}} = tan^{- 1} β - tan^{- 1} γ .

For the third term, the denominator

γ - α

is negative (since

α > γ

), so

cot^{- 1} {α + \frac{1 + α ^{2}}{γ - α}} = cot^{- 1} (\frac{α γ + 1}{γ - α}) = π + tan^{- 1} (\frac{γ - α}{1 + α γ}) = π + tan^{- 1} γ - tan^{- 1} α,

the extra

π

appearing because

cot^{- 1}

of a negative argument lands in

(\frac{π}{2}, π)

rather than matching the negative-angle branch of

tan^{- 1}

directly.

Adding all three terms:

(tan^{- 1} α - tan^{- 1} β) + (tan^{- 1} β - tan^{- 1} γ) + (π + tan^{- 1} γ - tan^{- 1} α) = π .

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option