Question

Prove ..Tan^-1(x)+tan^-1(y)=tan^-1(x+y/1-xy)

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Answer 1

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Answer 2

$\bf\ {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1} \frac{x + y}{1 - x} \\ \\ \tt\ l.h.s \sf\ \implies: {tan}^{ - 1}x + {tan}^{ - 1}y \\ \\ \\\sf\ let \: \: \red{{tan}^{ - 1}x = \theta } \\ \\ \tt\ \blue{ {tan}^{ - 1}y = \phi } \\ \\ \therefore \: tan( \theta + \phi) = \frac{(tan \theta + tan \phi)}{1 - tan \theta \: tan \phi} \\ \\ \sf\ \implies:tan( \theta + \phi) = \frac{x + y}{1 - xy} \\ \\ \sf\ \implies:( \theta + \phi) = {tan}^{ - 1}( \frac{x + y}{1 - xy}) \\ \\ \boxed{\sf\ \implies: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1}(\frac{x + y}{1 - xy}) }$