Evaluate the limitlim n tends to infinitytanA + 1/2 tan A/2 + 1/4 tan A/4 + ... 1/2n tan A/2n

Dear Eshita,

Ans:- Let F(A)=tan A+ 1/2 tanB+.............

So, F(A)dA= (tanA+1/2tanA/2+1/4tanA/4+.......................)dA

      ∫F(A)dA =∫(tanA+1/2tanA/2+.................)dA

                 =(ln secA+ln secA/2 +....................+ln secA/2^π) +C

                    =ln(secA secA/2 secA/4...............secA/2^n) +C

                        =ln(2^n/(cosA cosA/2 cosA/4..............)+C

                        =ln (2^n sinA/2^n/sin2A) +C

Again differentiating both sides we get,

F(A)=(1/2^n) cot(A/2^n)  - 2cot2A

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