3 years ago

Breaking information down into manageable pieces can help you:

Mathematics

1 answer:

scZoUnD [109]3 years ago

7 0

Answer no answers

Step-by-step explanation:

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Step-by-step explanation:

$\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )$

and $1+\tan ^2\theta =\sec^2\theta$

and $\cos \theta =\frac{1}{\sec \theta }$

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Applying this we get

$\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]$

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such that $\sin^2\theta +\cos^2\theta =1$

therefore

$\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002$

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