Question

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

Answer 1

Answer:1002

Explanation:

`\Rightarrow \sum_(k=1)^(2004)\left ( (1)/(1+\tan ^2\left ( (k\pi )/(2\cdot 2005)\right ))\right )`

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

and
`\cos \theta =(1)/(\sec \theta )`

`\Rightarrow \sum_(k=1)^(2004)\left ( \cos^2(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 ( (k\pi )/(2\cdot 2005)\right )+\cos^2\left ( ((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 ( (k\pi )/(2\cdot 2005)\right )+\cos^2\left ( ((2005-k)\pi )/(2\cdot 2005)\right )\right ]=1002`