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```				   Evaluate the limit
lim n tends to infinity
tanA + 1/2 tan A/2 + 1/4 tan A/4 + ... 1/2n tan A/2n
```

6 years ago

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```										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

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Eshita !!!

Regards,
SOUMYAJIT IIT_KHARAGPUR
```
6 years ago

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