If sinâ¡(θ/2)=a,cosâ¡(θ/2)=b, then (1+sinâ¡Î¸)(3sinâ¡Î¸+4cosâ¡Î¸+5)=
(a+b)2(a+3b)2
(a+b)2(3a+b)2
(aâb)2(aâ3b)2
(aâb)2(3aâb)2
(1+sinâ¡Î¸)(3sinâ¡Î¸+4cosâ¡Î¸+5)=(cosâ¡(θ/2)+sinâ¡(θ/2))2[6sinâ¡(θ/2)cosâ¡(θ/2)+42cos2â¡Î¸/2â1+5 =(a+b)26ab+8b2+a2+b2=(a+b)2(a+3b)2