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Question Number 201561 by Mingma last updated on 09/Dec/23

Answered by ajfour last updated on 11/Dec/23

Commented by ajfour last updated on 11/Dec/23

2θ+2φ=(π/2)  tan (θ+φ)=1  tan θ+tan φ=1−tan θtan φ      tan θ=(q/r)=(r/(p+a)) ⇒  pq+aq=r^2   tan φ=(p/r)=(r/(q+b)) ⇒ pq+bp=r^2   ⇒  ((tan θ)/(tan φ))=(q/p)   &  aq=bp  (subtracting)  hence   ((tan θ)/(tan φ))=(b/a)  2pq+aq+bp=2r^2      (adding)  2(rtan φ)(rtan θ)+r(atan θ+btan φ)         =2r^2   ⇒ (1/(2r))(atan θ+btan φ)=1−tan θtan φ  but   1−tan θtan φ=tan θ+tan φ  &  btan φ=atan θ    ⇒  (1/(2r))(2atan θ)=tan θ+(a/b)tan θ  ⇒   (1/r)=(1/a)+(1/b)

2θ+2ϕ=π2tan(θ+ϕ)=1tanθ+tanϕ=1tanθtanϕtanθ=qr=rp+apq+aq=r2tanϕ=pr=rq+bpq+bp=r2tanθtanϕ=qp&aq=bp(subtracting)hencetanθtanϕ=ba2pq+aq+bp=2r2(adding)2(rtanϕ)(rtanθ)+r(atanθ+btanϕ)=2r212r(atanθ+btanϕ)=1tanθtanϕbut1tanθtanϕ=tanθ+tanϕ&btanϕ=atanθ12r(2atanθ)=tanθ+abtanθ1r=1a+1b

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