If sin⁡(θ/2)=a,cos⁡(θ/2)=b, then (1+sin⁡θ)(3sin⁡θ+4cos⁡θ+5)=

(1+sin⁡θ)(3sin⁡θ+4cos⁡θ+5)=(cos⁡(θ/2)+sin⁡(θ/2))2[6sin⁡(θ/2)cos⁡(θ/2)+42cos2⁡θ/2−1+5 =(a+b)26ab+8b2+a2+b2=(a+b)2(a+3b)2