If cosα+cosβ=0=sinα+sinβ,

Solution:

Given: $\cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta$

$\therefore \cos \alpha+\cos \beta=0$

Squaring on both sides, we get

 $\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta=0$             ....(1)

Also,

$\sin \alpha+\sin \beta=0$

Squaring on both sides, we get

$\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta=0$           ...(2)

Subtracting (2) from (1), we get

$\left(\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta\right)-\left(\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta\right)=0$

 

$\Rightarrow \cos ^{2} \alpha-\sin ^{2} \alpha+\cos ^{2} \beta-\sin ^{2} \beta+2(\cos \alpha \cos \beta-\sin \alpha \sin \beta)=0$

$\Rightarrow \cos 2 \alpha+\cos 2 \beta+2 \cos (\alpha+\beta)=0$

 

$\Rightarrow \cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta)$