Question

tan^-1 x + tan^-1 y + tan^-1 = pi÷2 , then find the value of xy + yz + zx

Answer 1

${ \tan }^{ - 1} x + { \tan }^{ - 1} y + { \tan }^{ - 1} z = \frac{\pi}{2} \\ { \tan }^{ - 1} x + { \tan }^{ - 1} y = \frac{\pi}{2} - { \tan }^{ - 1} z \\ { \tan }^{ - 1} ( \frac{x + y}{1 - xy} ) = { \cot }^{ - 1} z = { \tan}^{ - 1} ( \frac{1}{z} ) \\ therefore \\ \frac{x + y}{1 - xy} = \frac{1}{z} \\ xz + yz = 1 - xy \\ xy + xz + yz = 1$