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Question Number 189941 by Rupesh123 last updated on 24/Mar/23

Answered by Frix last updated on 24/Mar/23

2arctan x +arctan y =arctan ((x^2 y−2x−y)/(x^2 +2xy−1)) answer is 0

2arctanx+arctany=arctanx2y−2x−yx2+2xy−1answeris0

Commented by Rupesh123 last updated on 24/Mar/23

Nice solution, sir!

Answered by CElcedricjunior last updated on 26/Mar/23

2.arctan(((b+c)/a))+arctan(((2a(b+c))/((b+c)^2 −a^2 )))=? tan(2𝛉+𝛃)=((tan2𝛉+tan𝛃)/(1−tan2𝛉tan𝛃)) =((((2tan𝛉)/(1−tan^2 𝛉))+tan𝛃)/(1−((2tan𝛉tan𝛃)/(1−tan^2 𝛉)))) =((2tan𝛉+tan𝛃−tan^2 𝛉×tan𝛃)/(1−tan^2 𝛉−2tan𝛉×tan𝛃)) =((2(((b+c)/a))+((2a(b+c))/((b+c)^2 −a^2 ))−(((b+c)/a))^2 (((2a(b+c))/((b+c)^2 −a^2 ))))/(1−(((b+c)^2 )/a^2 )−2(((b+c))/a)(((2a(b+c)))/([(b+c)^2 −a^2 ])))) =(((([2a(b+c)][(b+c)^2 −a^2 ]+2a^3 (b+c)−2a(b+c)^2 )/(a[(b+c)^2 −a^2 ])) )/((a^2 (b+c)^2 −a^4 −(b+c)^4 +a^2 (b+c)^2 −4a^2 (b+c)^2 )/(a^2 [(b+c)^2 −a^2 ]))) =((2a^2 (b+c)^2 [(b+c)−1))/(−a^4 −(b+c)^4 −2a^2 (b+c)^2 )) =((2a^2 (b+c)^2 [(b+c)^2 −1])/((a^2 +(b+c)^2 )^2 )) =

2.arctan(b+ca)+arctan(2a(b+c)(b+c)2−a2)=?tan(2θ+β)=tan2θ+tanβ1−tan2θtanβ=2tanθ1−tanθ2+tanβ1−2tanθtanβ1−tanθ2=2tanθ+tanβ−tanθ2×tanβ1−tanθ2−2tanθ×tanβ=2(b+ca)+2a(b+c)(b+c)2−a2−(b+ca)2(2a(b+c)(b+c)2−a2)1−(b+c)2a2−2(b+c)a(2a(b+c))[(b+c)2−a2]=[2a(b+c)][(b+c)2−a2]+2a3(b+c)−2a(b+c)2a[(b+c)2−a2]a2(b+c)2−a4−(b+c)4+a2(b+c)2−4a2(b+c)2a2[(b+c)2−a2]=2a2(b+c)2[(b+c)−1)−a4−(b+c)4−2a2(b+c)2=2a2(b+c)2[(b+c)2−1](a2+(b+c)2)2=

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