If x=2sin⁡θ1+cos⁡θ+sin⁡θ, then 1−cos⁡θ+sin⁡θ1+sin⁡θis equal to

Multiplying the numerator and denominator of x by 1−cos⁡θ+sin⁡θ, we getx=2sin⁡θ(1−cos⁡θ+sin⁡θ)(1+sin⁡θ)2−cos2⁡θ=2sin⁡θ(1−cos⁡θ+sin⁡θ)(1+sin⁡θ)2−1−sin2⁡θ=2sin⁡θ⋅(1−cos⁡θ+sin⁡θ)(1+sin⁡θ)(2sin⁡θ)=1−cos⁡θ+sin⁡θ1+sin⁡θ