If α>β>γ>0, then the expression

Question

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:

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

(2) \(3 \pi\)

(3) 0

(4) \(\pi\)
 

Answer 1

Correct option is (4) \(\pi\)   

\(\Rightarrow \cot ^{-1}\left(\frac{\alpha \beta+1}{\alpha-\beta}\right)+\cot ^{-1}\left(\frac{\beta \gamma+1}{\beta-\gamma}\right)+\cot ^{-1}\left(\frac{\alpha \gamma+1}{\gamma-\alpha}\right)\)

\(\Rightarrow \tan ^{-1}\left(\frac{\alpha-\beta}{1+\alpha \beta}\right)+\tan ^{-1}\left(\frac{\beta-\gamma}{1+\beta \gamma}\right)+\pi+\tan ^{-1}\left(\frac{\gamma-\alpha}{1+\gamma \alpha}\right)\)

\(\Rightarrow\left(\tan ^{-1} \alpha-\tan ^{-1} \beta\right)+\left(\tan ^{-1} \beta-\tan ^{-1} \gamma\right)+\left(\pi+\tan ^{-1} \gamma-\tan ^{-1} \alpha\right)\)

\(\Rightarrow \pi\)