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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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