If tan⁻¹x+tan⁻¹y+tan⁻¹z = π, then prove that x+y+z=xyz
Answers
Answered by
42
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]
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]
Answered by
42
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
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
Similar questions