If 0 , then the expression cot – 1 { + ( 1 + 2 ) } + cot – 1 { + ( 1 + 2 ) + cot – 1 {+ ( 1 + 2 ) is equal to [2025]

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Solution

Q.
If $α > β > γ > 0$ , 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: [2025]

1 $3 π$

2 0

3 $\frac{π}{2} - (α + β + γ)$

4 $π$

Ans.
(4)

$β + \frac{(1 + β^{2})}{α - β} = \frac{α β - β^{2} + 1 + β^{2}}{α - β} = \frac{α β + 1}{α - β}$

Similarly, $γ + \frac{(1 + γ^{2})}{β - γ} = \frac{β γ + 1}{β - γ} and α + \frac{(1 + α^{2})}{γ - α} = \frac{α γ + 1}{γ - α}$

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

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

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

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

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