Answer:1002
Step-by-step explanation:
[tex]\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )[/tex]
and [tex]1+\tan ^2\theta =\sec^2\theta [/tex]
and [tex]\cos \theta =\frac{1}{\sec \theta }[/tex]
[tex]\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )[/tex]
as [tex]\cos ^2\theta =\cos ^2(\pi -\theta )[/tex]
Applying this we get
[tex]\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 ][/tex]
every [tex]\theta [/tex]there exist [tex]\pi -\theta[/tex]
such that [tex]\sin^2\theta +\cos^2\theta =1[/tex]
therefore
[tex]\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[/tex]
