If tan^{-1} x + tan^{-1} y = frac{pi}{4},then | vedclass

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(B) Given the equation: $	an^{-1} x + 	an^{-1} y = \frac{\pi}{4}$.Using the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x + y}{1 - x

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$x + y + xy = 1$

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(B) Given the equation: $	an^{-1} x + 	an^{-1} y = \frac{\pi}{4}$.Using the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x + y}{1 - xy} ight)$,we get:$	an^{-1} \left( \frac{x + y}{1 - xy} ight) = \frac{\pi}{4}$.Taking the tangent of both sides:$\frac{x + y}{1 - xy} = 	an \left( \frac{\pi}{4} ight)$.Since $	an \left( \frac{\pi}{4} ight) = 1$,we have:$\frac{x + y}{1 - xy} = 1$.Multiplying both sides by $(1 - xy)$,we get:$x + y = 1 - xy$.Rearranging the terms,we obtain:$x + y + xy = 1$.

If $\tan^{-1} x + \tan^{-1} y = \frac{\pi}{4}$,then:

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