Question

${ \tan}^{ - 1} x \: + \: { \tan}^{ - 1} y \: + \: { \tan }^{ - 1} z \: = \pi$
then show that
x + y + z = xyz​

Answer 1

Question :-

${ \tan}^{ - 1} x \: + \: { \tan}^{ - 1} y \: + \: { \tan }^{ - 1} z \: = \pi$

then show that

x + y + z = xyz

Step by step explanation:-

Formula used :-

$\red{ { \tan}^{ - 1} x + { \tan }^{ - 1} y + { \tan}^{ - 1} z \: } \\ \\ = \green{ { \tan }^{ - 1} \bigg( \frac{x + y + z - xyz}{1 - xy - yz - zx} \bigg)}$

Solution :-

According to the question,

${ \tan}^{ - 1} x + { \tan }^{ - 1} y + { \tan}^{ - 1} z \: \: = \pi$

Applied the given formula →

${ \tan }^{ - 1} \bigg( \frac{x + y + z - xyz}{1 - xy - yz - zx} \bigg) = \pi \\ \\ \bigg( \frac{x + y + z - xyz}{1 - xy - yz - zx} \bigg) \: = \tan( \pi) \\ \\ \because \: \tan(\pi) = 0 \\ \\ \therefore \: \bigg( \frac{x + y + z - xyz}{1 - xy - yz - zx} \bigg) \: = 0 \\ \\ \implies \: x + y + z - xyz = 0 \\ \\ \implies \: \boxed{ x + y + z = xyz}$

Hence proved.