Trignometric identities question

taking Lcm and solve we get     sin²Φ(1-sin²θ)² + sin⁴θcos²Φ = sin²Φ (1-sin²Φ )

further solving in terma of sinΦ and sinθ       we get    sin⁴θ + sin⁴Φ  = 2sin²Φ sin²θ                                           which gives (sin²θ - sin²Φ)² =0   =====> sin θ  =  sinΦ   ===> θ =  Φ 

so putting θ =  Φ in second eqn. we get  sin²θ + cos²θ = 1    which is true