The value of 2 cos

$2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$

Using identity:-  2 cos x cos y = cos (x + y) + cos (x – y)

$=\cos \left(\frac{9 \pi}{13}+\frac{\pi}{13}\right)+\cos \left(\frac{9 \pi}{13}-\frac{\pi}{13}\right)+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$

$=\cos \left(\frac{10 \pi}{13}\right)+\cos \left(\frac{8 \pi}{13}\right)+\cos \left(\frac{3 \pi}{13}\right)+\cos \left(\frac{5 \pi}{13}\right)$

$=\cos \left(\frac{10 \pi}{13}\right)+\cos \left(\frac{3 \pi}{13}\right)+\cos \left(\frac{8 \pi}{13}\right)+\cos \left(\frac{5 \pi}{13}\right)$

$\left[\right.$ using identity, $\left.\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\right]$

$=2 \cos \left(\frac{\frac{10 \pi}{13}+\frac{3 \pi}{13}}{2}\right) \cos \left(\frac{\frac{10 \pi}{13}-\frac{3 \pi}{13}}{2}\right)+2 \cos \left(\frac{\frac{8 \pi}{13}+\frac{5 \pi}{13}}{2}\right) \cos \left(\frac{\frac{8 \pi}{13}-\frac{5 \pi}{13}}{2}\right)$

$=2 \cos \left(\frac{\frac{13 \pi}{13}}{2}\right) \cos \left(\frac{\frac{7 \pi}{13}}{2}\right)+2 \cos \left(\frac{\frac{13 \pi}{13}}{2}\right) \cos \left(\frac{\frac{4 \pi}{13}}{2}\right)$

$=2 \cos \frac{\pi}{2} \cos \frac{7 \pi}{26}+2 \cos \frac{\pi}{2} \cos \frac{4 \pi}{26}$

$=2 \cos \frac{\pi}{2}\left(\cos \frac{7 \pi}{26}+\frac{4 \pi}{26}\right)$

$=2 \times 0\left(\cos \frac{7 \pi}{26}+\cos \frac{4 \pi}{26}\right)=0$

Hence, the answer is 0.