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Question Number 144031 by mathdanisur last updated on 20/Jun/21

Answered by mindispower last updated on 20/Jun/21

$$\frac{1}{sin\left(2^k\right)}+\frac{1}{tg\left(2^k\right)}=\frac{1+cos\left(2k\right)}{sin\left(2k\right)}=\frac{1}{tg\left(2^{k-1}\right)}$$$$\frac{1}{sin\left(2^k\right)}=\frac{1}{tg\left(2^{k-1}\right)}-\frac{1}{tg\left(2^k\right)}$$$$\underset{k=1}{\sum ^n}\frac{1}{sin\left(2^k\right)}=\frac{1}{tg\left(1\right)}--\frac{1}{tg\left(2^n\right)}$$

Commented by mathdanisur last updated on 22/Jun/21

$$ThanksSircool$$

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