If {tan ^{ - 1}}x - {tan ^{ - 1}}y = {tan ^{ | vedclass

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(C) We are given the equation: ${	an ^{ - 1}}x - {	an ^{ - 1}}y = {	an ^{ - 1}}A$. Using the standard trigonometric identity for the difference of

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$\frac{x - y}{1 + xy}$

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(C) We are given the equation: ${	an ^{ - 1}}x - {	an ^{ - 1}}y = {	an ^{ - 1}}A$. Using the standard trigonometric identity for the difference of inverse tangents: ${	an ^{ - 1}}x - {	an ^{ - 1}}y = {	an ^{ - 1}}\left( \frac{x - y}{1 + xy} ight)$. Comparing this with the given equation: ${	an ^{ - 1}}\left( \frac{x - y}{1 + xy} ight) = {	an ^{ - 1}}A$. Therefore,by equating the arguments,we get: $A = \frac{x - y}{1 + xy}$.

If ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}A,$ then $A$ is equal to:

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