Question

Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]

Answer 1

Answer:1002

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

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

as $\cos ^2\theta =\cos ^2(\pi -\theta )$

Applying this we get

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

every $\theta$there exist $\pi -\theta$

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

therefore

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