If cos⁡x−sin⁡α cot⁡β sin⁡x=cos⁡α, then the value of tan (x/2) is

cos⁡x−sin⁡αcot⁡βsin⁡x=cos⁡α⇒1−tan2⁡(x/2)1+tan2⁡(x/2)−sin⁡αcot⁡β2tan⁡(x/2)1+tan2⁡(x/2)=cos⁡α⇒tan2⁡x2(1+cos⁡α)+2sin⁡αcot⁡βtan⁡x2−(1−cos⁡α)=0⇒tan2⁡x2+2sin⁡αcot⁡β1+cos⁡αtan⁡x2−1−cos⁡α1+cos⁡α=0⇒tan2⁡x2+2tan⁡α2cot⁡βtan⁡x2−tan2⁡α2=0⇒tan2⁡x2+2tan⁡α2⋅12cot⁡β2−tan⁡β2tan⁡x2−tan2⁡α2=0⇒tan⁡x2+cot⁡β2tan⁡α2tan⁡x2−tan⁡β2tan⁡α2=0⇒tan⁡x2=−tan⁡α2cot⁡β2or tan⁡x2=tan⁡α2tan⁡β2