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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=π2tan(θ+ϕ)=1tanθ+tanϕ=1−tanθtanϕtanθ=qr=rp+a⇒pq+aq=r2tanϕ=pr=rq+b⇒pq+bp=r2⇒tanθtanϕ=qp&aq=bp(subtracting)hencetanθtanϕ=ba2pq+aq+bp=2r2(adding)2(rtanϕ)(rtanθ)+r(atanθ+btanϕ)=2r2⇒12r(atanθ+btanϕ)=1−tanθtanϕbut1−tanθtanϕ=tanθ+tanϕ&btanϕ=atanθ⇒12r(2atanθ)=tanθ+abtanθ⇒1r=1a+1b
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