Question

if x+y+z=xyz then proove that 2x/1-x^2+'2y/1-y^2+2z/1-z^2 =2x/1-x^2*2y/1-y^2*2z/1-z^2......

Answer 1

♥heya friend !!!.✔..

let that x=tanA,y=tanB,z=tanC

given ,,x+y+z=xyz

since,tanA+tanB+tanC=tanA*tanB
..

or,tanA+tanB+tanC=tanA*tanB*tanC.

or,tanA+tanB=tanA*tanB*tanC-tanC

or,tanA+tanB=-tanC(1-tanA.tanB)

or,tanA+tanB
-----------------------=-tanC
1-tanA*tanB

or,tan*(A+B)=tan(180°-C)

A+B=180°-C

A+B+C=180°

or,2A+2B+2C=360°

or,tan(2A+2B)=tan(360°-2C) 【multiplied by tan】

tan2A+Tan2B/1-tan2A*tan2B=-tan2C

or,tan2A+tan2B/1-tanA*tanB=-tan2A

or,tan2A+tan2B+tan2C=tan2A*tan2B*tan2C

or,2tanA/1-tan^2A+2tan2B/1-tan^2B+2tanC/1-tanC=2tanA/1-tan^2A×2tan2B/1-tan^2B×2tanC/1-tan^2C

or,2x/1-x^2+2y/1-y^2+2z/1-z^2=2x/1-x^2*2y/1-y^2*2z/1-z^2...【if we putting the value of x,y,z)

.hope it help you...♥☺☺☺☺

.your BRAINLY STAR Rajukumar .....☺☺☺

Answer 2

let that x=tanA,y=tanB,z=tanC

given ,,x+y+z=xyz

since,tanA+tanB+tanC=tanA*tanB

..

or,tanA+tanB+tanC=tanA*tanB*tanC.

or,tanA+tanB=tanA*tanB*tanC-tanC

or,tanA+tanB=-tanC(1-tanA.tanB)

or,tanA+tanB

-----------------------=-tanC

1-tanA*tanB

or,tan*(A+B)=tan(180°-C)

A+B=180°-C

A+B+C=180°

or,2A+2B+2C=360°

or,tan(2A+2B)=tan(360°-2C) 【multiplied by tan】

tan2A+Tan2B/1-tan2A*tan2B=-tan2C

or,tan2A+tan2B/1-tanA*tanB=-tan2A

or,tan2A+tan2B+tan2C=tan2A*tan2B*tan2C

or,2tanA/1-tan^2A+2tan2B/1-tan^2B+2tanC/1-tanC=2tanA/1-tan^2A×2tan2B/1-tan^2B×2tanC/1-tan^2C

or,2x/1-x^2+2y/1-y^2+2z/1-z^2=2x/1-x^2*2y/1-y^2*2z/1-z^2...【if we putting the value of x,y,z)