Question

If tan⁻¹x+tan⁻¹y+tan⁻¹z = π, then prove that x+y+z=xyz

Answer 1

Given, tan^-1x + tan^-1y + tan^-1z = π

tan^-1x + tan^-1y = π - tan^-1z

we know, tan^-1P + tan^-1Q = tan^-1(P + Q)/(1 - PQ) , where PQ < 1

so, tan^-1x + tan^-1y = tan^-1(x + y)(1 - xy)

so, tan^-1(x + y)/(1 - xy) = π - tan^-1z

(x + y)/(1 - xy) = tan(π - tan^-z)

we know, tan(π - ∅) = - tan∅

so, (x + y)/(1 - xy) =- tan(tan^-1(z))

=> (x + y)/(1 - xy) = -z

=> (x + y) = (1 - xy)(-z) = -z + xyz

=> x + y + z = xyz [ hence proved]

Answer 2

HELLO DEAR,



GIVEN:-

tan-¹x + tan-¹y + tan-¹z = π

tan-¹x + tan-¹y = π - tan-¹z

we know, tan-¹a + tan-¹b = tan-¹(a + b)/(1 - ab)

therefore,

tan-¹(x + y)(1 - xy) = π - tan-¹z

taking tangent both side,

(x + y)/(1 - xy) = tan(π - tan-¹z)

[ as, tan(π - A) = - tanA ]

so, (x + y)/(1 - xy) = - tan{tan-¹(z)}

=> (x + y)/(1 - xy) = -z

=> (x + y) = (1 - xy)(-z)

=> x + y = -z + xyz

=> x + y + z = xyz



I HOPE ITS HELP YOU DEAR,
THANKS