It is assumed x=tanθ, then we have atanθ+bsecθ=c asinθ+b=ccosθ cosθ−asinθ=b Let, a2+c2=r2 c=rcosα,a=rsinα tanα=
a
c
cosα×cosθ−sinα×sinθ=
b
r
Thus, cos(α+θ)=
b
r
α+θ=±cos−1
b
r
Let, a+θ be the positive solution and a+ϕ the negative solution, where y=tanϕ α+ϕ=−(α+θ) −2α=θ+ϕ tan(−2α)=tan(θ+ϕ) On expanding the above expression we get.
−2tanα
1−tan2α
=tan(θ+ϕ) Substitute the values in above expression.