The value of ∑n=1∞ tan⁡θ2n2n−1cos⁡θ2n−1

Sn=∑r=1n tan⁡θ2r2r−1cos⁡θ2r−1=∑r=1n sin⁡θ2r2r−1cos⁡θ2rcos⁡θ2r−1=∑r=1n 2sin2⁡θ2r2r−12sin⁡θ2rcos⁡θ2rcos⁡θ2r−1=∑r=1n 1−cos⁡θ2r−12r−1sin⁡θ2r−1cos⁡θ2r−1=∑r=1n 12r−2sin⁡θ2r−2−12r−1sin⁡θ2r−1=2sin⁡2θ−12n−1sin⁡θ2n−1∴ S∞=2sin⁡2θ−limn→∞ θ2n−1⋅1θsin⁡θ2n−1=2sin⁡2θ−1θ