If (x−a) cos⁡ θ+y sin⁡ θ=(x−a) cos⁡ ϕ+y sin⁡ ϕ=a and tan⁡ (θ/2)−tan⁡ (ϕ/2)=2b, then

Let tan⁡ (θ/2)=α and tan⁡(ϕ/2)=β, so that α−β=2b.Also, cos⁡θ=1−tan2⁡(θ/2)1+tan2⁡(θ/2)=1−α21+α2and sin⁡θ=2tan⁡(θ/2)1+tan2⁡(θ/2)=2α1+α2Similarly,      cos⁡ϕ=1−β21+β2 and sin⁡ϕ=2β1+β2Therefore, we have from the given relations(x−a)1−α21+α2+y2α1+α2=a⇒ xα2−2yα+2a−x=0Similarly xβ2−2yβ+2a−x=0We see that α and β are roots of the equationxz2−2yz+2a−x=0,So that α+β=2y/x and αβ=(2a−x)/x.Now, from (α+β)2=(α−β)2+4αβ, we get⇒ 2yx2=(2b)2+4(2a−x)x⇒ y2=2ax−1−b2x2Also, from α+β=2y/x and α−β=2b, we getα=y/x+b and β=y/x−b⇒ tan⁡θ2=1x(y+bx) and tan⁡ϕ2=1x(y−bx)