If alpha beta gamma 0,then the expressi | vedclass

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(D) Let the given expression be $S = \cot ^{-1}\left\{\beta+\frac{1+\beta^2}{\alpha-\beta}\right\}+\cot ^{-1}\left\{\gamma+\frac{1+\gamma^2}{\beta-\ga

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(D) Let the given expression be $S = \cot ^{-1}\left\{\beta+\frac{1+\beta^2}{\alpha-\beta}ight\}+\cot ^{-1}\left\{\gamma+\frac{1+\gamma^2}{\beta-\gamma}ight\}+\cot ^{-1}\left\{\alpha+\frac{1+\alpha^2}{\gamma-\alpha}ight\}$.Since $\cot^{-1}(x) = 	an^{-1}(1/x)$ for $x > 0$,we rewrite the terms:$S = 	an^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}ight) + 	an^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}ight) + 	an^{-1}\left(\frac{\gamma-\alpha}{1+\gamma\alpha}ight)$.Using the identity $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$:$S = (	an^{-1}\alpha - 	an^{-1}\beta) + (	an^{-1}\beta - 	an^{-1}\gamma) + 	an^{-1}\left(\frac{\gamma-\alpha}{1+\gamma\alpha}ight)$.Since $\gamma  -1$ and $x Thus,$S = (	an^{-1}\alpha - 	an^{-1}\beta) + (	an^{-1}\beta - 	an^{-1}\gamma) + (	an^{-1}\gamma - 	an^{-1}\alpha - \pi) = -\pi$. Wait,checking the sign convention: $\cot^{-1}(x) = 	an^{-1}(1/x)$ for $x>0$. The expression simplifies to $\pi$.

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

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