Let 3cos⁡θ+5cos2⁡θ+7cos3⁡θ+…∞=−12, where cos⁡θ∣<1. The value of 1+2sin2⁡θ22 is equal to

S=3cos⁡θ+5cos2⁡θ+7cos3⁡θ+⋯∞⇒cos⁡θS=3cos2⁡θ+5cos3⁡θ+⋯∞ So,(1−cos⁡θ)S=3cos⁡θ+2cos2⁡θ+cos3⁡θ+⋯∞⇒ (1−cos⁡θ)S=3cos⁡θ+2cos2⁡θ1−cos⁡θ⇒(1−cos⁡θ)2−12=3cos⁡θ(1−cos⁡θ)+2cos2⁡θ⇒−cos2⁡θ+2cos⁡θ−1=6cos⁡θ−6cos2⁡θ+4cos2⁡θ⇒cos2⁡θ−4cos⁡θ−1=0⇒cos⁡θ=2±5⇒cos⁡θ=2−51+2sin2⁡θ22=(1+1−cos⁡θ)2=(2−(2−5))2=5